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MATHEMATICAL TEXTS 

Edited by PERCEY F. SMITH, Pn.D 

Professor of Mathematics in the Sheffield Scientific School 
of Yale University 



Elements of the Differential and Integral Calculus 
(Revised Edition) 

By W. A. GRANVILLE, PH.D. 
Elements of Analytic Geometry 

By P. F. SMITH and A. S. GALE, PH.D. 
New Analytic Geometry 

By P. F. SMITH and A. S. GALE, PH.D. 
Introduction to Analytic Geometry . 

By P. F. SMITH and A. S. GALE, PH.D. 
Advanced Algebra 

By H. E. HAWKES, PH.D. 
Text-Book on the Strength of Materials (Revised Edition) 

By S. E. SLOCUM, PH.D., and E. L. HANCOCK, M.Sc. 
Problems in the Strength of Materials 

By WILLIAM KENT SHEPARD, PH.D. 
Plane and Spherical Trigonometry and Four-Place 
. Tables of Logarithms 

By W. A. GRANVILLE, PH.D. 
Plane and Spherical Trigonometry 

By W. A. GRANVILLE, PH.D. 
Plane Trigonometry and Four-Place Tables of Logarithms 

By W. A. GRANVILLE, PH.D. 
Four-Place Tables of Logarithms 

By W. A. GRANVILLE, PH.D. 
Theoretical Mechanics 

By P. F. SMITH and W. R. LONGLEY, PH.D. 
First Course in Algebra 

By H. E. HAWKES, PH.D., WILLIAM A. LUBY, A.B., 

and FRANK C. TOUTON, Pn.B. 

Second Course in Algebra 

By H. E. HAWKES, PH.D., WILLIAM A. LUBY, A.B., 
and FRANK C. TOUTON, Pn.B. 

Elementary Analysis 

By P. F. SMITH and W. A. GRANVILLE, Pn.D 



TEXT-BOOK ON THE 
STRENGTH OF MATERIALS 



BY 

S. E. SLOCUM, B.E., PH.D. 

PROFESSOR OF APPLIED MATHEMATICS IN THE UNIVERSITY OF CINCINNATI 



E. L. HANCOCK, M.S. 

PROFESSOR OF APPLIED MECHANICS IN WORCESTER POLYTECHNIC INSTITUTE 



REVISED EDITION 



GINN AND COMPANY 

BOSTON NEW YORK CHICAGO LONDON 







COPYRIGHT, 1906, 1911, BY 
S. E. SLOCUM AND E. L. HANCOCK 



ALL RIGHTS RESERVED 
811.10 



gfce 



GINN AND COMPANY PRO- 
PRIETORS BOSTON U.S.A. 



PREFACE 



Five years of extensive use of this book, since the appearance of 
the first edition, have brought to the authors from various sources 
numerous suggestions relating to its improvement. In particular the 
authors wish to acknowledge their indebtedness to Professor Irving P. 
Church of Cornell University and to Professor George R Chatburn of 
the University of Nebraska for their unfailing interest and frequent 
valuable suggestions. 

To utilize the material so obtained, the text has been thoroughly 
revised. In making this revision the aim of the authors has been 
twofold: first, to keep the text abreast of the most recent practical 
developments of the subject; and second, to simplify the method 
of presentation so as to make the subject easily intelligible to the 
average technical student of junior grade, as well as to lessen the 
work of instruction. 

Besides correcting the errors inevitable to a first edition, special 
attention has been given to amplifying the explanation wherever ex- 
perience in using the book as a text has indicated it to be desirable. 
This applies especially to the articles on Poisson's ratio, the theorem 
of three moments, the calculation of the stress in curved members, 
the relation of Guest's and Eankine's formulas to the design of shafts 
subjected to combined stresses, etc. 

Considerable new material has also been added. In Part I a set 
of tables has been placed at the beginning of the volume to facilitate 
numerical calculations. Other important additions are articles on 
the design of reenforced concrete beams, shrinkage and forced fits, the 
design of eccentrically loaded columns, the design and efficiency of 
riveted joints, the general theory of the torsion of springs, practical 
formulas for the collapse of tubes, and an extension of the method of 
least work to a wide variety of practical problems. This last includes 



265507 



vi STRENGTH OF MATERIALS 

the derivation and application of the Fraenkel formula for the bending 
deflection of beams, and also a simple general formula for the shearing 
deflection of beams, never before published. 

Nearly one hundred and fifty original problems have also been 
added to Part I. These problems are designed not merely to provide 
numerical exercises on the text, but have been selected throughout 
with the specific purpose of emphasizing the practical importance of 
the subject and extending the range of its application as widely as 
possible. Many of them are practical shop problems brought up by 
students in the cooperative engineering course at the University of 
Cincinnati. 

In Part II the recent advances in the manufacture of steel have 
been given special attention, including the properties of vanadium 
steel, manganese steel, and high-speed steel. Reenforced concrete 
has also received a more adequate treatment, and the chapter on this 
subject has been thoroughly revised and modernized. The chapter on 
timber has also received an equally thorough revision, and considerable 
material on preservative processes has been added. 

In both the first edition and the present revision, Part I, covering 
the analytical treatment of the subject, is the work of S. E. Slocum, 
and Part II, presenting the experimental or laboratory side, is the 
work of E. L Hancock. TH AUTHORg 



CONTENTS 

PART I MECHANICS OF MATERIALS 

CHAPTER I 
ELASTIC PROPERTIES OF MATERIALS 

PAGES 

Introductory. Subject-matter of the strength of materials. 
Stress, strain, and deformation. Tension, compression, and shear. 

Unit stress. Unit deformation. Strain diagrams. Hooke's 
law and Young's modulus. Poisson's ratio. Ultimate strength. 
Elastic law. Classification of materials. Time effect. Fatigue 
of metals. Hardening effects of overstraining. Fragility. 
Initial internal stress. Annealing. Temperature stresses. Effect 
of length, diameter, and form of cross section. Factor of safety. 

Work done in producing strain 1~19 

CHAPTER II 

FUNDAMENTAL RELATIONS BETWEEN STRESS AND 
DEFORMATION 

Relations between the stress components. Planar strain. Stress 
in different directions. Maximum normal stress. Principal 
stresses. Maximum shear. Linear strain. Stress ellipse. 
Simple shear. Coefficient of expansion. Modulus of elasticity of 
shear. Relation between the elastic constants. Measure of strain. 

Combined bending and torsion 20-34 

CHAPTER III 
ANALYSIS OF STRESS IN BEAMS 

System of equivalent forces. Common theory of flexure. Ber- 
noulli's assumption. Curvature due to bending moment. Conse- 
quence of Bernoulli's assumption. Result of straight-line law. 
Moment of inertia. Moment of resistance. Section modulus. 
Theorems on the moment of inertia. Graphical method of find- 
ing the moment of inertia. Moment of inertia of non-homogeneous 
sections. Inertia ellipse. Vertical reactions and shear. Maxi- 
mum bending moment. Bending moment and shear diagrams. 

vii 



viii STRENGTH OF MATERIALS 

PAGES 

Relation between shear and bending moment. Designing of beams. 
Distribution of shear over rectangular cross section. Distribution 
of shear over circular cross section. Cases in which shear is of 
especial importance. Oblique loading. Eccentric loading. 
Antipole and antipolar. Core section. Application to concrete 
and masonry construction. Calculation of pure bending strain by 
means of the core section. Stress trajectories. Materials which 
do not conform to Hooke's law. Design of ree'nf orced concrete beams 35-80 

CHAPTER IV 
FLEXURE OF BEAMS 

Elastic curve. Limitation to Bernoulli's assumption. Effect of 
shear on the elastic curve. Built-in beams. Continuous beams. 

Theorem of three moments. Work of deformation Impact 

and resilience. Influence line for bending moment. Influence 
line for shear. Maxwell's theorem. Influence line for reactions. 

Castigliano's theorem. Application of Castigliano's theorem to 
continuous beams. Principle of least work. General formula for 
flexural deflection. General formula for shearing deflection . . . 81-119 

CHAPTER V 
COLUMNS AND STRUTS 

Nature of compressive stress. Euler's theory of long columns. 

Columns with one or both ends fixed. Independent proof of 
formulas for fixed ends. Modification of Euler's formula. Ran- 
kine's formula. Values of the empirical constants in Rankine's 
formula. Johnson's parabolic formula. Johnson's straight-line 
formula. Cooper's modification of Johnson's straight-line formula. 

Beams of considerable depth. Eccentrically loaded columns . . 120-137 

CHAPTER VI 
TORSION 

Circular shafts. Maximum stress in circular shafts. Angle of 
twist in circular shafts. Power transmitted by circular shafts. 
Combined bending and torsion. Resilience of circular shafts. 
Non-circular shafts. Elliptical shaft. Rectangular and square 
shafts. Triangular shafts. Angle of twist for shafts in general. 

Helical springs. General theory of spiral springs 138-153 



CONTENTS ix 

CHAPTER VII 
SPHERES AND CYLINDERS UNDER UNIFORM PRESSURE 

Hoop stress. Hoop tension in hollow sphere. Hoop tension in 
hollow circular cylinder. Longitudinal stress in hollow circular 
cylinder. Differential equation of elastic curve for circular cylinder. 
Crushing strength of hollow circular cylinder. Thick cylinders ; 
Lamp's formulas. Maximum stress in thick cylinder under uniform 
internal pressure. Bursting pressure for thick cylinder. Maxi- 
mum stress in thick cylinder under uniform external pressure. 
Thick cylinders built up of concentric tubes. Practical formulas for 
the collapse Of tubes under external pressure. Shrinkage and forced 
fits. Riveted joints 154-178 

CHAPTER VIII 
FLAT PLATES 

Theory of flat plates. Maximum stress in homogeneous circular 
plate under uniform load. Maximum stress in homogeneous circu- 
lar plate under -concentrated load. Dangerous section of elliptical 
plate. Maximum stress in homogeneous elliptical plate under uni- 
form load. Maximum stress in homogeneous square plate under 
uniform load. Maximum stress in homogeneous rectangular plate 
under uniform load. Xon-homogeneous plates ; concrete-steel floor 
panels 179-190 

CHAPTER IX 
CURVED PIECES : HOOKS, LINKS, AND SPRINGS 

Erroneous analysis of hooks and links. Bending strain in curved 
piece. Simplification of formula for unit stress. Curved piece of 
rectangular cross section. Effect of sharp curvature on bending 
strength. Maximum moment in circular piece. Plane spiral 
springs 191-206 

CHAPTER X 

ARCHES AND ARCHED RIBS 

I. GRAPHICAL ANALYSIS OF FORCES 

Composition of forces. Equilibrium polygon. Application of 
equilibrium polygon to determining reactions. Equilibrium poly- 
gon through two given points. Equilibrium polygon through three 



X STRENGTH OF MATERIALS 

PAGES 

given points. Application of equilibrium polygon to calculation of 

stresses Relation of equilibrium polygon to bending moment 

diagram / . . . 207-216 

II. CONCRETE AND MASONRY ARCHES 

Definitions and construction of arches. Load line. Linear arch. 
Conditions for stability. Maximum compressive stress. Loca- 
tion of the linear arch : Moseley's theory. Application of the 
principle of least work. Winkler's criterion for stability. Em- 
pirical formulas. Designing of arches. Stability of abutments. 
Oblique projection of arch 216-230 

III. ARCHED RIBS 

Stress in arched ribs. Three-hinged arched rib. Two-hinged 
arched rib. Second method of calculating the pole distance. 
Graphical determination of the linear arch. Temperature stresses 
in two-hinged arched rib. Continuous arched rib fixed at both 
ends. Graphical determination of the linear arch for continuous 
arched rib. Temperature stresses in continuous arched rib ... 230-242 

CHAPTER XI 
FOUNDATIONS AND RETAINING WALLS 

Bearing power of soils. Angle of repose and coefficient of fric- 
tion. Bearing power of piles. Ordinary foundations. Column 
footings. Maximum earth pressure against retaining walls. 
Stability of retaining walls. Thickness of retaining walls . . . 243-262 



PART II PHYSICAL PROPERTIES OF MATERIALS 

CHAPTER XII 
IRON AND STEEL 

Introductory. Tension tests. Compression tests. Flexure 
tests. Method of holding tension specimens. Behavior of iron 
and steel in tension. Effect of overstrain on wrought iron and mild 
steel. Relative strength of large and small test pieces. Strength 
of iron and steel at high temperatures. Character and appearance 



CONTENTS 



XI 



of the fracture. Measurement of extension, compression, and de- 
flection. Torsion tests. Form of torsion test specimen. Torsion 
as a test of shear. Shearing tests. Impact tests. Cold bending 
tests. Cast iron. Strain diagram for cast iron. Cast iron in 
flexure. Cast iron in shear. Cast-iron columns. Malleable cast- 
ings. Specifications for cast iron. Wrought iron and steel. 
Manufacture of steel. Composition of steel. Steel castings. 

Modulus of elasticity of steel and wrought iron. Standard form 

of test specimens. Specifications for wrought iron and steel . . 265-296 

CHAPTER XIII 
LIME, CEMENT, AND CONCRETE 

Quicklime. Cement. Cement tests. Test of soundness. 
Test of fineness. Test of time of setting.' Test of tensile strength. 

Speed of application of load. Compression tests. Standard 
specifications for cement. Concrete. Mixing of concrete. Tests 
of concrete. Modulus of elasticity of concrete. Cinder concrete. 

Concrete building blocks. Effect of temperature on the strength 

of concrete .... . 297-312 



CHAPTER XIV 
REENFORCED CONCRETE 

Object of reenforcement. Corrosion of the metal reinforcement. 
Adhesion of the concrete to the reenforcement. Area of the metal 
reenf orcement. Position of the neutral axis in reenf orced concrete 
beams. Strength of reenf orced concrete beams. Linear variation 
of stress. Bond between steel and concrete. Strength of T-beams. 
Shear at the neutral axis . 313-325 



CHAPTER XV 
BRICK AND BUILDING STONE 

Limestone. Sandstone. Compression tests of stone. Trans- 
verse tests of stone. Abrasion tests of stone. Absorption tests 
of stone. Brick and brickwork. Compression tests of brick. 
Modulus of elasticity of brick. Transverse tests of brick. Rattler 
test of brick. Absorption test of brick 326-335 






xii STRENGTH OF MATERIALS 

PAGES 

CHAPTER XVI 
TIMBER 

Structure of timber. Annual rings. Heartwood and sapwood. 
Effect of moisture. Strength of timber. Compression tests. 
Flexure tests. Shearing tests. Indentation tests. Tension tests. 
European tests of timber. Tests made for the tenth census. 
Tests made by the Bureau of Forestry. Recent work of the United 
States Forest Service. Treated timber. Strength of treated 
timber 336-355 

CHAPTER XVII 
ROPE, WIRE, AND BELTING 

Wire. Wire rope. Testing of rope, wire, and belting. Strength 
of wire rope. Strength of manila rope. Strength of leather and 
rubber belting 356-363 

ANSWERS TO PROBLEMS 365-366 

INDEX . 367-372 



NOTATION 

The references are to articles. 

A, B, C, Constant coefficients, 49, 85. 

C\, Cj, etc., Constants of integration, 67, 85. 

D, Deflection, 67, 107. 

E, E 8 , E c , Young's modulus, 8. 

F, FI, -F 2 , Area, 5. 

G, Modulus of shear, 33. 
H, Horse power, 99. 

7, I x , J a , Ip, etc., Moment of inertia, 43. 

Ji k , Influence numbers, 77. 

K, Coefficient of cubical expansion, 32. 

L, * Coefficient of linear expansion, 19. 

M, MQ, MI, M 2 , External moment, 43. 

N, Statical moment, 47. 

P, P', P h , etc., Concentrated force, 5, 85, 86, 171. 

Q, Resultant shear, 53. 

P T> D / Reactions of abutments, 50, 172. 

\ Resistance of soil, 168. 
S, Section modulus, 45, 170. 

T, Temperature change, 19. 

F, Volume, 32. 

W, Work, 73, 81. 



f Semi-axis of ellipse, 49, 59, 103. 
\ Radius of shaft, 97. 
( Semi-axis of ellipse, 49, 59, 103. 
6, J Radius of shaft, 97. 

[Breadth, 43, 104. 

c, Distance, 46, 47, 52, 170. 
C Symbol of differentiation. 

d, J Diameter of shaft, 99. 

[Distance, 52, 67, 100, 168. 

e, Distance of extreme fiber from neutral axis, 43. 
/, Empirical constant, 89. 

( Empirical constant, 89. 

\Factor of safety, 172. 
h, Height, depth, 66. 

( Coefficient of friction, 167. 
fc, \ Constant, 115, 127, 132. 

[Number, 229. 



xiv STRENGTH OF MATERIALS 

I, Length, distance, 6, 47, 49, 85. 

m, Poisson's constant, 9. 

f Abstract number, 92, 99, 170. 

\ Ratio, 66. 

P,P\,P',Px,ete; Unit normal stress, '5, 23, 25. 
p e , Equivalent normal stress, 35. 

<7> (lit <l'i <lx, etc., Unit shear, 5, 23, 25. 

( Radius, 46, 56, 96. 

\Ratio, 161, 227. 

s, Unit deformation, 6. 

, t x , t a , etc., Radius of gyration, 46, 49. 

f Curvilinear coordinate, 113. 

I Bond, 229. 

M t , Ultimate tensile strength, 117, 169. 

w c , Ultimate compressive strength, 148, 149. 

f Unit load, 51. 

[Weight per cubic foot, 171. 
x, y, 2, Variables. 

x, y, z, Coordinates of center of gravity, 42. 



a, Angle, 25, 46, 171. 

/3, Angle, 67, 75, 171. 

5, e, Empirical constants, 91. 

r, Angle, 171. 

17, Correction coefficient, 65. 

f Angle of twist, 96. 

\Angle, 172. 

/c, Ratio between tensile and shearing strength, 57. 

X, Arbitrary integer, 26, 85. 

yu, Constant, 99. 

v, Empirical constant, 11, 92. 

TT, Ratio of circumference to diameter, 

p, Radius of curvature, 67, 113. 

0-, Empirical constant, 11, 92. 

S, Symbol of summation, 25. 

0, Angle of shear, 33, 96. 

w, Angle of repose, 167. 



TABLES OF PHYSICAL AND MATHEMATICAL 

CONSTANTS 

I. AVERAGE VALUES OF PHYSICAL CONSTANTS 

II. PROPERTIES OF VARIOUS SECTIONS 

III. PROPERTIES OF STANDARD I-BEAMS 

IV. PROPERTIES OF STANDARD CHANNELS 
V. PROPERTIES OF STANDARD ANGLES 

VI. MOMENTS OF INERTIA AND SECTION MODULI : RECTANGULAR CROSS 

SECTION 
VII. MOMENTS OF INERTIA AND SECTION MODULI : CIRCULAR CROSS 

SECTION 

VIII. FOUR-PLACE LOGARITHMS OF NUMBERS 
IX. CONVERSION OF LOGARITHMS 
X. FUNCTIONS OF ANGLES 
XL BENDING MOMENT AND SHEAR DIAGRAMS 



TABLES 



xvii 



"! a 

"OH 



ulus 
hear 
lus o 
dity) 



Hit 2 



M 
of 
o 
Ki 



Young's 
Modulus o 
Elasticity 



. 
Slil 



ii < 

5H 5 



flfff 

5SoQ| 



0202 



Ultimate 
Tensile 
Strength 



gf g 



of 



(M (M O 3tJ 

1 1 i 1 

0000 



I 



8 g 



a s g 



co co 



8 s 



S S 



a 



5 "ic 2 

1 1 1 1 

K OQ ^ O 



? 8 . . 

i 3 In 

S o S 



XV111 



STRENGTH OF MATERIALS 



2. POISSON'S RATIO 



MATERIAL 


AVERAGE 
VALUES OF 
1 
m 


Steel hard 


295 


" structural 


299 




.277 


Brass . 


357 


Copper 


340 


Lead . . 


.375 


Zinc 


205 







3. FACTORS OF SAFETY 



MATERIAL 


STEADY 
STRESS : 
BUILDINGS, 

ETC. 


VARYING 
STRESS : 
BRIDGES, 

ETC. 


REPEATED OR 
REVERSED 
STRESS : 
MACHINES 


Steel hard 


5 


8 


15 


" structural 
Iron wrought 


4 
4 


6 
6 


10 
10 


" cast 


6 


10 


20 


Timber 


8 


10 


15 


Brick and stone 


15 


25 


30 











The only rational method of determining the factor of safety is to choose it 
sufficiently large to bring the working stress well within the elastic limit (see 
Article 14). 



H a 

M 



H 

ES 



a 



o x 



-^ 



O I <N 
II 
S 




xix 



fa O 
H 



s 



* |(N 
O |rH 




SIS! 



C<> ICO 

|| 

s 




<M ico 
II 
H 




1 





i 

ft 




SI* 



i 

ts ITU 




< -X 

H 



> 



ICO 

> 

CO 




ICO 

> s 




?si 




xxii 




ttl 



h- 




xxiii 



XXIV 



STRENGTH OF MATERIALS 



TABLE III 
PROPERTIES OF STANDARD I-BEAMS 



DEPTH 

OF 

BEAM 


WEIGHT 

PER 

FOOT 


AREA 

OF 

SECTION 


THICK- 
NESS OF 
WEB 


WIDTH 

OF 

FLANGE 


MOMENT 

OF 

INERTIA 
Axis 1-1 


SECTION 
MODU- 
LUS 
Axis 1-1 


RADIUS 

OF 

GYRA- 
TION 
Axis 1-1 


MOMENT 

OF 

INERTIA 
Axis 2-2 


RADIUS 

OF 

GYRA- 
TION 
Axis 2-2 


d 




A 


t 


b 


I 


S 


r 


i' 


r' 


Inches 


Pounds 


Sq.hwhes 


Inches 


Inches 


Inches * 


Inches 3 


Inches 


Inches * 


Inches 


3 


55 


1.63 


.17 


2.33 


2.5 


1.7 


1.23 


.46 


53 




6.5 


1.91 


.26 


2.42 


2.7 


1.8 


1.19 


53 


52 


" 


7.5 


2.21 


.36 


252 


2.9 


1.9 


1.15 


.60 


52 


4 


7.5 


2.21 


.19 


2.66 


6.0 


3.0 


1.64 


.77 


.59 





8.5 


2.50 


.26 


2.73 


6.4 


3.2 


1.59 


.85 


.58 





9.5 


2.79 


.34 


2.81 


6.7 


3.4 


1.54 


.93 


58 


" 


10.5 


3.09 


.41 


2.88 


7.1 


3.6 


152 


1.01 


57 


5 


9.75 


2.87 


.21 


3.00 


12.1 


4.8 


2.05 


1.23 


.65 




12.25 


3.60 


.36 


3.15 


13.6 


5.4 


1.94 


1.45 


.63 


" 


14.75 


4.34 


.50 


3.29 


15.1 


6.1 


1.87 


1.70 


.63 


6 


12.25 


3.61 


.23 


3.33 


21.8 


7.3 


2.46 


1.85 


.72 


d 


14.75 


4.34 


.35 


3.45 


24.0 


8.0 


2.35 


2.09 


.69 


" 


17.25 


5.07 


.47 


357 


26.2 


8.7 


2.27 


2.36 


.68 


7 


15.0 


4.42 


.25 


3.66 


36.2 


10.4 


2.86 


2.67 


.78 


" 


17.5 


5.15 


.35 


3.76 


39.2 


11.2 


2.76 


2.94 


.76 


" 


20.0 


5.88 


.46 


3.87 


42.2 


12.1 


2.68 


3.24 


.74 


8 


17.75 


5.33 


.27 


4.00 


56.9 


14.2 


3.27 


3.78 


.84 





20.25 


5.96 


.35 


4.08 


60.2 


15.0 


3.18 


4.04 


.82 


" 


22.75 


6.69 


.44 


4.17 


64.1 


16.0 


3.10 


4.36 


.81 


" 


25.25 


7.43 


.53 


4.26 


68.0 


17.0 


3.03 


4.71 


.80 


9 


21.0 


6.31 


.29 


4.33 


84.9 


18.9 


3.67 


5.16 


.90 


" 


25.0 


7.35 


.41 


4.45 


91.9 


20.4 


3.54 


5.65 


.88 


" 


30.0 


8.82 


57 


4.61 


101.9 


22.6 


3.40 


6.42 


.85 


" 


35.0 


10.29 


.73 


4.77 


111.8 


24.8 


3.30 


7.31 


.84 


10 


25.0 


7.37 


.31 


4.66 


122.1 


24.4 


4.07 


6.89 


.97 





30.0 


8.82 


.45 


4.80 


134.2 


26.8 


3.90 


7.65 


.93 


< 


35.0 


10.29 


.60 


4.95 


146.4 


29.3 


3.77 


8.52 


.91 


" 


40.0 


11.76 


.75 


5.10 


158.7 


31.7 


3.67 


9.50 


.90 


12 


32.5 


9.26 


.35 


5.00 


215.8 


36.0 


4.83 


9.50 


1.01 


" 


35.0 


10.29 


.44 


5.09 


228.3 


38.0 


4.71 


10.07 


.99 


" 


40.0 


11.76 


56 


5.21 


245.9 


41.0 


457 


10.95 


.96 


15 


42.0 


12.48 


.41 


550 


441.8 


58.9 


5.95 


14.62 


1.08 


" 


45.0 


13.24 


.46 


555 


455.8 


60.8 


5.87 


15.09 


1.07 


" 


50.0 


14.71 


56 


5.65 


483.4 


645 


5.73 


16.04 


1.04 


" 


55.0 


16.18 


.66 


5.75 


511.0 


68.1 


5.62 


17.06 


1.03 


" 


60.0 


17.65 


.75 


5.84 


538.6 


71.8 


552 


18.17 


1.01 


18 


55.0 


15.93 


.46 


6.00 


795.6 


88.4 


7.07 


21.19 


1.15 


" 


60.0 


17.65 


56 


6.10 


841.8 


93.5 


6.91 


22.38 


1.13 


" 


65.0 


19.12 


.64 


6.18 


881.5 


97.9 


6.79 


23.47 


1.11 


" 


70.0 


20.59 


.72 


6.26 


921.2 


102.4 


6.69 


24.62 


1.09 


20 


65.0 


19.08 


.50 


6.25 


1169.5 


117.0 


7.83 


27.86 


1.21 





70.0 


20.59 


58 


6.33 


1219.8 


122.0 


7.70 


29.04 


1.19 


" 


75.0 


22.06 


.65 


6.40 


1268.8 


126.9 


758 


30.25 


1.17 


24 


80.0 


23.32 


50 


7.00 


2087.2 


173.9 


9.46 


42.86 


1.36 


" 


85.0 


25.00 


57 


7.07 


2167.8 


180.7 


9.31 


44.35 


1.33 


" 


90.0 


26.47 


.63 


7.13 


2238.4 


1865 


9.20 


45.70 


1.31 


" 


95.0 


27.94 


.69 


7.19 


2309.0 


192.4 


9.09 


77.10 


1.30 


" 


100.0 


29.41 


.75 


7.25 


2379.6 


198.3 


8.99 


4855 


1.28 



TABLES 



XXV 



TABLE IV 
PROPERTIES OF STANDARD CHANNELS 

1 





M 





fe 




V 


to 


i 


O1 


03 
J 





J w 


DEPTH OF 
CHANNEL 


WEIGHT PEI 
FOOT 


REA OF SECT! 


THICKNESS o 
WEB 


WIDTH OF 
FLANGE 


MOMENT OF 

fERTIA AXIS 


SCTION'MODU 

Axis 1-1 


ADIUS OF GY 
TION AXIS 1- 


MOMENT OF 
STERTIA AXIS 


:CTION MODI- 
AXIS 2-2 


ADIUS OF GYl 
TION AXIS 2- 


DISTANCE 01 
ENTER OF GR. 
Y FROM OUTS 
OF WEB 






< 






M 


02 


W 


1-1 


HH 

cc 


Pq 


^S 


d 




A 


t 


b 


I 


S 


r 


i' 


S' 


r' 


X 


Inches 


Pounds 


Sq.In. 


Inches 


Inches 


Inches* 


Inches 5 


Inches 


Inches 


Inches* 


Inches 


Inches 


3 


4.00 


1.19 


.17 


1.41 


1.6 


1.1 


1.17 


'.20 


.21 


.41 


.44 


ii 


5.00 


1.47 


.26 


1.50 


1.8 


1.2 


1.12 


.25 


.24 


.41 


.44 


" 


6.00 


1.76 


.36 


1.60 


2.1 


1.4 


1.08 


.31 


!27 


.42 


.46 


4 


5.25 


1.55 


.18 


1.58 


3.8 


1.9 


1.56 


.32 


.29 


.45 


.46 


Cl 


6.25 


1.84 


.25 


1.65 


4.2 


2.1 


1.51 


.38 


.32 


.45 


.46 


" 


7.25 


2.13 


.33 


1.73 


4.6 


2.3 


1.46 


.44 


.35 


.46 


.46 


5 


6.50 


1.95 


.19 


1.75 


7.4 


3.0 


1.95 


.48 


.38 


.50 


.49 




9.00 


2.65 


.33 


1.89 


8.9 


3.5 


1.83 


.64 


.45 


.49 


.48 


" 


11.50 


3.38 


.48 


2.04 


10.4 


4.2 


1.75 


.82 


.54 


.49 


.51 


6 


8.00 


2.38 


.20 


1.92 


13.0 


4.3 


2.34 


.70 


.50 


.54 


.52 


" 


10.50 


3.09 


.32 


2.04 


15.1 


5.0 


2.21 


.88 


.57 


.53 


.50 


a 


13.00 


3.82 


.44 


2.16 


17.3 


5.8 


2.13 


1.07 


.65 


.53 


.52 


" 


15.50 


4.56 


.56 


2.28 


19.5 


6.5 


2.07 


1.28 


.74 


.53 


.55 


7 


9.75 


2.85 


.21 


2.09 


21.1 


6.0 


2.72 


.98 


.63 


.59 


.55 




12.25 


3.60 


.32 


2.20 


24.2 


6.9 


2.59 


1.19 


.71 


.57 


.53 


d 


14.75 


4.34 


.42 


2.30 


27.2 


7.8 


2.50 


1.40 


.79 


.57 


.53 


" 


17.25 


5.07- 


.53 


2.41 


30.2 


8.6 


2.44 


1.62 


.87 


.56 


.55 


" 


19.75 


5.81 


.63 


2.51 


33.2 


9.5 


2.39 


1.85 


.96 


.56 


.58 


8 


11.25 


3.35 


.22 


2.26 


32.3 


8.1 


3.10 


1.33 


.79 


.63 


.58 


" 


13.75 


4.04 


.31 


2.35 


36.0 


9.0 


2.98 


1.55 


.87 


.62 


.56 


ii 


16.25 


4.78 


]40 


2.44 


39.9 


10.0 


2.89 


1.78 


.95 


.61 


.56 


" 


18.75 


5.51 


.49 


2.53 


43.8 


11.0 


2.82 


2.01 


1.02 


.60 


.57 


" 


21.25 


6.25 


.58 


2.62 


47.8 


11.9 


2.76 


2.25 


1.11 


.60 


.59 


9 


13.25 


3.89 


.23 


2.43 


47.3 


10.5 


3.49 


1.77 


.97 


.67 


.61 




15.00 


4.41 


.29 


2.49 


50.9 


11.3 


3.40 


1.95 


1.03 


.66 


.59 


a 


20.00 


5.88 


.45 


2.65 


60.8 


13.5 


3.21 


2.45 


1.19 


.65 


.58 


" 


25.00 


7.35 


.61 


2.81 


70.7 


15.7 


3.10 


2.98 


1.36 


.64 


.62 


10 


15.00 


4.46 


.24 


2.60 


66.9 


13.4 


3.87 


2.30 


1.17 


.72 


.64 




20.00 


5.88 


.38 


2.74 


78.7 


15.7 


3.66 


2.85 


1.34 


.70 


.61 


" 


25.00 


7.35 


.53 


2.89 


91.0 


18.2 


3.52 


3.40 


1.50 


.68 


.62 


ii 


30.00 


8.82 


.68 


3.04 


103.2 


20.6 


3.42 


3.99 


1.67 


.67 


.65 


" 


35.00 


10.29 


.82 


3.18 


115.5 


23.1 


3.35 


4.66 


1.87 


.67 


.69 


12 


20.50 


6.03 


.28 


2.94 


128.1 


21.4 


4.61 


3.91 


1.75 


.81 


.70 


ii 


25.00 


7.35 


.39 


3.05 


144.0 


24.0 


4.43 


4.53 


1.91 


.78 


.68 


" 


30.00 


8.82 


.51 


3.17 


161.6 


26.9 


4.28 


5.21 


2.09 


.77 


.68 


ii 


35.00 


10.29 


.64 


3.30 


179.3 


29.9 


4.17 


5.90 


2.27 


.76 


.69 


" 


40.00 


11.76 


.76 


3.42 


196.9 


32.8 


4.09 


6.63 


2.46 


.75 


.72 


15 


33.00 


9.90 


.40 


3.40 


312.6 


41.7 


5.62 


8.23 


3.16 


.91 


.79 


" 


35.00 


10.29 


.43 


3.43 


319.9 


42.7 


5.57 


8.48 


3.22 


.91 


.79 


" 


40.00 


11.76 


.52 


3.52 


347.5 


46.3 


5.44 


9.39 


3.43 


.89 


.78 


< 


45.00 


13.24 


.62 


3.62 


375.1 


50.0 


5.32 


10.29 


3.63 


.88 


.79 





50.00 


14.71 


.72 


3.72 


402.7 


53.7 


5.23 


11.22 


3.85 


.87 


.80 




55.00 


16.18 


.82 


3.82 


430.2 


57.4 


5.16 


12.19 


4.07 


.87 


.82 



XXVI 



STRENGTH OF MATERIALS 



TABLE V 
PROPERTIES OF STANDARD ANGLES, EQUAL LEGS 





\ 
i 


\ 


\ 








i 


v \ i 




\ t 


] 




DIMENSIONS 


THICKNESS 


WEIGHT PER FOOT 


AREA OF SECTION 


DISTANCE OF CENTER 
OF GRAVITY FROM 
BACK OF FLANGE 


MOMENT OF INERTIA 
Axis 1-1 


SECTION MODULUS 
Axis 1-1 


RADIUS OF GYRA- 
TION Axis 1-1 


DISTANCE OF CENTER 
OF GRAVITY FROM 
EXTERNAL APEX ON 
LINE INCLINED AT 
45 TO FLANGE 


LEAST MOMENT OF 
INERTIA Axis 2-2 


SECTION MODULUS 
Axis 2-2 


LEAST RADIUS OF 
GYRATION Axis 2-2 


Inches 


Inches 


founds 


$q. In. 


Inches 


Inches* 


Inches 3 


Inches 


Inches 


Inches* 


Inches 3 


Inches 


fx j 


* 


\ 


58 


.17 


.23 


.009 


.017 


22 


33 


.004 


.011 


.14 


1 x 1 


j 





.80 


.23 


.30 


.022 


.031 


.30 


.42 


.009 


.021 


.19 




! 




1.49 


.44 


M 


.037 


.056 


.29 


.48 


.016 


.034 


.19 


I 1 x I 1 


, 





1.02 


.30 


.36 


.044 


.049 


.38 


51 


.018 


.035 


.24 


" 


! 


\ 


1.91 


56 


.40 


.077 


.091 


.37 


57 


.033 


.057 


.24 


I 5 x I' 


^ 





2.34 


.69 


.47 


.14 


.134 


.45 


.66 


.058 


.088 


.29 




\ 


! 


3.35 


.98 


51 


.19 


.188 


.44 


.72 


.082 


.114 


.29 


1| x \l 


: 




2.77 


.81 


53 


.23 


.19 


53 


.75 


.094 


.13 


.34 


" 


, 


! 


3.98 


1.17 


57 


.31 


.26 


51 


.81 


.133 


.16 


.34 


2 x 


i 




3.19 


.94 


59 


.35 


.25 


.61 


.84 


.14 


.17 


.39 


" 


' 


! 


4.62 


1.36 


.64 


.48 


.35 


59 


.90 


.20 


.22 


.39 


2 x 2 





; 


4.0 


1.19 


.72 


.70 


.39 


.77 


1.01 


.29 


.28 


.49 


" 


'. 




5.9 


1.73 


.76 


.98 


57 


.75 


1.08 


.41 


.38 


.48 


" 


i 


i 


7.7 


2.25 


.81 


1.23 


.72 


.74 


1.14 


52 


.46 


.48 


3x3 






4.9 


1.44 


.84 


1.24 


58 


.93 


1.19 


.50 


.42 


59 


" 






7.2 


2.11 


.89 


1.76 


.83 


.91 


1.26 


.72 


57 


.58 


i> 






9.4 


2.75 


.93 


2.22 


1.07 


.90 


1.32 


.92 


.70 


58 


" 






11.4 


3.36 


.98 


2.62 


1.30 


.88 


1.38 


1.12 


.81 


58 


3 x 3 






8.4 


2.48 


1.01 


2.87 


1.15 


1.07 


1.43 


1.16 


.81 


.68 


" 






11.1 


3.25 


1.06 


3.64 


1.49 


1.06 


150 


150 


1.00 


.68 


(i 








3.98 


1.10 


4.33 


1.81 


1.04 


156 


1.82 


1.17 


.68 


" 






15.9 


4.69 


1.15 


4.96 


2.11 


1.03 


1.62 


2.13 


1.31 


.67 


4x4 






9.7 


2.86 


1.14 


4.36 


152 


1.23 


1.61 


1.77 


1.10 


.79 








12.8 


3.75 


1.18 


556 


1.97 


1.22 


1.67 


2.28 


1.36 


.78 


i< 






15.7 


4.61 


1.23 


6.66 


2.40 


1.20 


1.74 


2.76 


1.59 


.77 


" 





: 


185 


5.44 


1.27 


7.66 


2.81 


1.19 


1.80 


3.23 


1.80 


.77 


6x6 





, 


19.6 


5.75 


1.68 


19.91 


4.61 


1.86 


2.38 


8.04 


3.37 


1.18 








24.2 


7.11 


1.73 


24.16 


5.66 


1.84 


2.45 


9.81 


4.01 


1.17 


n 




' 


28.7 


8.44 


1.78 


28.15 


6.66 


1.83 


251 


11.52 


459 


1.17 


11 


' 




33.1 


9.73 


1.82 


31.92 


7.63 


1.81 


257 


13.17 


5.12 


1.16 



TABLES 



XXVll 



TABLE V 
PROPERTIES OF STANDARD ANGLES, UNEQUAL LEGS 












a 


i 






K 


d 






DIMENSIONS 


THICKNESS 


WEIGHT PER FOOT 


AREA OF SECTION 


DISTANCE OF CENTE 
OF GRAVITY FROM 
BACK OF LONGER 
FLANGE 


MOMENT OF INERTI. 
Axis 1-1 


SECTION MODULUS 
Axis 1-1 


RADIUS OF GYRA- 
TION Axis 1-1 


DISTANCE OF CENTE 
OF GRAVITY FROM 
BACK OF SHORTER 
FLANGE 


MOMENT OF INERTI. 
Axis 2-2 


SECTION MODULUS 
Axis 2-2 


RADIUS OF GYRA- 
TION Axis 2-2 


Inches 


Inches 


I'ounds 


Sq.In. 


Inches 


Inches* 


Inches 3 


Inches 


Inches 


Inches* 


Inches 


Inches 


2$ x 2 


I 


3.6 


1.06 


.54 


.37 


.25 


59 


.79 


.65 


.38 


.78 





1 


5.3 


1.55 


.58 


51 


.36 


58 


.83 


.91 


.55 


.77 


" 




6.8 


2.00 


.63 


.64 


.46 


56 


.88 


1.14 


.70 


.75 


3 x 2 


i 


4.5 


1.31 


.66 


.74 


.40 


.75 


.91 


1.17 


56 


.95 


n 


1 


6.5 


1.92 


.71 


1.04 


.58 


.74 


.96 


1.66 


.81 


.93 


" 




8.5 


2.50 


.75 


1.30 


.74 


.72 


1.00 


2.08 


1.04 


.91 


3 x 2 


i 


4.9 


1.44 


.61 


.78 


.41 


.74 


1.11 


1.80 


.75 


1.12 


" 


i 


7.2 


2.11 


.66 


1.09 


59 


.72 


1.16 


2.56 


1.09 


1.10 


" 


i 


9.4 


2.75 


.70 


1.36 


.76 


.70 


1.20 


3.24 


1.41 


1.09 


" 


1 


11.4 


3.36 


.75 


1.61 


.92 


.69 


1.25 


3.85 


1.71 


1.07 


3$ x 3 


g 


7.8 


2.30 


.83 


1.85 


.85 


.90 


1.08 


2.72 


1.13 


1.09 





i 


10.2 


3.00 


.88 


2.33 


1.10 


.88 


1.13 


3.45 


1.45 


1.07 





1 


12.5 


3.67 


.92 


2.76 


1.33 


.87 


1.17 


4.11 


1.76 


1.06 


" 


1 


14.7 


4.31 


.96 


3.15 


1.54 


.85 


1.21 


4.70 


2.05 


1.04 


4x3 


I 


8.5 


2.48 


.78 


1.92 


.87 


.88 


1.28 


3.96 


1.46 


1.26 


" 


i 


11.1 


3.25 


.83 


2.42 


1.12 


.86 


1.33 


5.05 


1.89 


1.25 


" 


| 


13.6 


3.98 


.87 


2.87 


1.35 


.85 


1.37 


6.03 


2.30 


1.23 


" 


| 


15.9 


4.69 


.92 


3.28 


157 


.84 


1.42 


6.93 


2.68 


1.22 


5 X3 


I 


9.7 


2.86 


.70 


2.04 


.89 


.84 


1.70 


7.37 


2.24 


1.61 


" 


i 


12.8 


3.75 


.75 


2.58 


1.15 


.83 


1.75 


9.45 


2.91 


1.59 


" 





15.7 


4.61 


.80 


3.06 


1.39 


.82 


1.80 


11.37 


3.55 


157 


" 




18.5 


5.44 


.84 


3.51 


1.62 


.80 


1.84 


13.15 


4.16 


155 


5 x 3 


3 


10.4 


3.05 


.86 


3.18 


1.21 


1.02 


1.61 


7.78 


2.29 


1.60 


" 


1 


13.6 


4.00 


.91 


4.05 


156 


1.01 


1.66 


9.99 


2.99 


1.58 


" 


5 


16.7 


4.92 


.95 


4.83 


1.90 


.99 


1.70 


12.03 


3.65 


1.56 





f 


19.8 


5.81 


1.00 


5.55 


2.22 


.98 


1.75 


13.92 


4.28 


1.55 


" 


I 


22.7 


6.67 


1.04 


6.21 


252 


.96 


1.79 


15.67 


4.88 


153 


6 x 3J 


g 


11.6 


3.42 


.79 


3.34 


1.23 


.99 


2.04 


12.86 


3.24 


1.94 


" 


1 


15.3 


4.50 


.83 


4.25 


1.59 


.97 


2.08 


16.59 


4.24 


1.92 


" 


| 


18.9 


5.55 


.88 


5.08 


1.94 


.96 


2.13 


20.08 


5.19 


1.90 


< 


3 


22.3 


6.56 


.93 


5.84 


2.27 


.94 


2.18 


23.34 


6.10 


1.89 


" 


| 


25.7 


7.55 


.97 


6.55 


259 


.93 


2.22 


26.39 


6.98 


1.87 


6x4 


| 


12.3 


3.61 


.94 


4.90 


1.60 


1.17 


1.94 


13.47 


3.32 


1.93 


" 


1 


16.2 


4.75 


.99 


6.27 


2.08 


1.15 


1.99 


17.40 


4.33 


1.91 


K 


6 


19.9 


5.86 


1.03 


7.52 


2.54 


1.13 


2.03 


21.07 


5.31 


1.90 








23.6 


6.94 


1.08 


8.68 


2.97 


1.12 


2.08 


2451 


6.25 


1.88 


" 


1 


27.2 


7.98 


1.12 


9.75 


3.39 


1.11 


2.12 


27.73 


7.15 


1.86 



xxvm 



STRENGTH OF MATERIALS 



TABLE VI 

MOMENTS OF INERTIA AND SECTION MODULI : RECTANGULAR 
CROSS SECTION 



BREADTH | 

* 


HEIGHT 
h 


MOMENT 

OF 

INERTIA 

, bhs 
12 


SECTION 
MODULUS 

<, bh* 
6 


BREADTH 1 
b 


HEIGHT 
h \ 


MOMENT 

OF 

INERTIA 

r bh* 
12 


SECTION 
MODULUS 

,. 


BREADTH 
b 


HEIGHT 


MOMENT 

OF 

INERTIA 

, 6^3 

12 


SECTION 
MODULUS 

o 6/' 2 
~~G~ 


1 


I 

2 
3 
4 
5 

6 
7 
8 
9 
10 
11 
12 

2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 


.0833 
.66 
2.25 
5.33 
10.42 
18 
28.58 
42.66 
60.75 
83.33 
110.92 
144 


.166 
.66 
1.5 
2.66 
4.16 
6 
8.16 
10.66 
13.5 
16.66 
20.16 
24 


4 

5 


4 

5 
6 
7 
8 
9 
10 
11 
12 


21.33 
41.66 
72 
114.33 
170.66 
243 
333.33 
443.66 
576 


10.66 
16.66 
24 
32.66 
42.66 
54 
66.66 
80.66 
96 


8 
9 


8 
9 
10 
11 
12 
13 
14 
15 
16 


341.33 

486 
666.66 
887.33 
1152 
1464.66 
1829.33 
2250 
2730.66 


85.33 
108 
133.33 
161.33 
192 
225.33 
261.33 
300 
341.33 


5 
6 
7 
8 
9 
10 
11 
12 


52.08 
90 
142.92 
213.33 
303.75 
416.66 
554.58 
720 


20.83 
30 
40.83 
53.33 
67.5 
83.33 
100.83 
120 


9 
10 
11 
12 
13 
14 
15 
16 
17 
18 


546.75 
750 
998.25 
1296 
1647.75 
2058 
2531.25 
3072 
3684.75 
4374 


121.5 
150 
181.5 
216 
253.5 
294 
337.5 
384 
433.5 
486 


2 


1.33 
4.5 
10.66 
20.83 
36 
57.16 
85.33 
121.5 
166.66 
221.85 
288 


1.33 
3 
5.33 
8.33 
12 
16.33 
21.33 
27 
33.33 
40.33 
48 


6 


6 
7 
8 
9 
10 
11 
12 


108 
171.5 
256 
364.5 
500 
665.5 
864 


36 
49 
64 
81 
100 
121 
144 


10 


10 
11 
12 
.13 
14 
15 
16 
17 
18 
19 
20 


833.33 
1109.16 
1440 
1830.83 
2286.66 
2810 
3413.33 
4094.17 
4860 
5715.83 
6666.66 


166.66 
201.66 
240 
281.66 
326.66 
375 
426.66 
481.66 
540 
601.66 
666.66 


3 


3 
4 

5 
6 
7 
8 
9 
10 
11 
12 


6.75 
16 
31.25 
54 
85.75 
128 
182.25 
250 
332.75 
432 


4.5 

8 
12.5 
18 
24.5 
32 
40.5 
50 
60.5 
72 


7 


7 
8 
9 
10 
11 
12 
13 
14 


200.08 
298.66 
425.25 
583.33 
776.42 
1008 
1281.58 
1600.66 


57.16 
74.66 
94.5 
116.66 
141.16 
168 
197.16 
228.66 



TABLES 



XXIX 



TABLE VII 

MOMENTS OF INERTIA AND SECTION MODULI : CIRCULAR 
CROSS SECTION 



DlAM- 1 
ETER 


MOMENT 

OF 

INERTIA 


SECTION 
MODULUS 


DIAM- 
ETER 


MOMENT 

OF 

INERTIA 


SECTION 
MODULUS 


DIAM- 
ETER 


MOMENT 

OF 

INERTIA 


SECTION 
MODULUS 


1 

2 
3 
4 
5 
6 
7 
8 
9 
10 


.0491 
.7854 
3.976 
12.57 
30.68 
63.62 
117.9 
201.1 
322.1 
490.9 


.0982 
.7854 
2.651 
6.283 
12.27 
21.21 
33.67 
50.27 
71.57 
98.17 


35 
36 
37 
38 
39 
40 


73,662 
82,448 
91,998 
102,354 
113,561 
125,664 


4,209 
4,580 
4,973 

5,387 
5,824 
6,283 


69 
70 


1,112,660 
1,178,588 


32,251 
33,674 


71 

72 
73 
74 
75 

76 
77 
78 
79 
80 


1,247,393 
1,319,167 
1,393,995 
1,471,963 
1,553,156 
1,637,662 
1,725,571 
1,816,972 
1,911,967 
2,010,619 


35,138 
36,644 
38,192 
39,783 
41,417 
43,096 
44,820 
46,589 
48,404 
50,265 


41 
42 
43 
44 
45 
46 
47 
48 
49 
50 


138,709 
152,745 
167,820 
183,984 
201,289 
219,787 
239,531 
260,576 
282,979 
306,796 


6,766 
7,274 
7,806 
8,363 
8,946 
9,556 
10,193 
10,857 
11,550 
12,270 


11 
12 
13 
14 
15 
16 
17 
18 
19 
20 


718.7 
1,018 
1,402 
1,886 
2,485 
3,217 
4,100 
5,153 
6,397 
7,854 


130.7 
169.6 
215.7 
269.4 
331.3 
402.1 
482.3 
572.6 
673.4 
785.4 


81 
82 
83 
84 
85 
86 
87 
88 
89 
90 


2,113,051 
2,219,347 
2,329,605 
2,443,920 
2,562,392 
2,685,120 
2,812,205 
2,943,748 
3,079,853 
3,220,623 


52,174 
54,130 
56,135 
58,189 
60,292 
62,445 
64,648 
66,903 
69,210 
71,569 


51 
52 
53 
54 
55 
56 
57 
58 
59 
60 


332,086 
358,908 
387,323 
417,393 
449,180 
482,750 
518,166 
555,497 
594,810 
636,172 


13,023 
13,804 
14,616 
15,459 
16,334 
17,241 
18,181 
19,155 
20,163 
21,206 


21 
22 
23 
24 
25 
26 
27 
28 
29 
30 


9,547 
11,499 
13,737 
16,286 
19,175 
22,432 
26,087 
30,172 
34,719 
39,761 


909.2 
1,045 
1,194 
1,357 
1,534 
1,726 
1,932 
2,155 
2,394 
2,651 


91 
92 
93 
94 
95 
96 
97 
98 
99 
100 


3,366,165 
3,516,586 
3,671,992 
3,832,492 
3,998,198 
4,169,220 
4,345,671 
4,527,664 
4,715,315 
4,908,727 


73,982 
76,448 
78,968 
81,542 
84,173 
86,859 
89,601 
92,401 
95,259 
98,175 


61 
62 
63 
64 
65 
66 
67 
68 


679,651 
725,332 
773,272 
823,550 
876,240 
931,420 
989,166 
1,049,556 


22,284 
23,398 
24,548 
25,736 
26,961 
28,225 
29,527 
30,869 


31 
32 
33 
34 


45,333 
51,472 
58,214 
65,597 


2,925 
3,217 

3,528 
3,859 



TABLE VIII 
FOUR-PLACE LOGARITHMS OF NUMBERS 



1 





1 


2 


3 


4 


5 


6 


7 


8 


9 





0000 


0000 


3010 


4771 


6021 


6990 


7782 


8451 


9031 


9542 


1 


0000 


0414 


0792 


1139 


1461 


1761 


2041 


2304 


2553 


2788 


2 


3010 


3222 


3424 


3617 


3802 


3979 


4150 


4314 


4472 


4624 


3 


4771 


4914 


5051 


5185 


5315 


5441 


5563 


5682 


5798 


5911 


4 


6021 


6128 


6232 


6335 


6435 


6532 


6628 


6721 


6812 


6902 


5 


6990 


7076 


7160 


7243 


7324 


7404 


7482 


7559 


7634 


7709 


6 


7782 


7853 


7924 


7993 


8062 


8129 


8195 


8261 


8325 


8388 


7 


8451 


8513 


8573 


8633 


8692 


8751 


8808 


8865 


8921 


8976 


8 


9031 


9085 


9138 


9191 


9243 


9294 


9345 


9395 


9445 


9494 


9 


9542 


9590 


9638 


9685 


9731 


9777 


9823 


9868 


9912 


9956 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


6729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


50 





1 


2 


3 


4 


5 


6 


7 


8 


9 



TABLES 



xxxi 



50 





1 


2 


3 


4 


5 


6 


7 


8 


9 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8803 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754. 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


QS 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


100 





1 


2 


3 


4 


5 


6 


7 


8 


9 



XXX11 



STRENGTH OF MATERIALS 



TABLE IX 
CONVERSION OF LOGARITHMS 

REDUCTION OF COMMON LOGARITHMS TO NATURAL LOGARITHMS 

Rule for using Table. Divide the given common logarithm into periods of two 
digits and take from the table the corresponding numbers, having regard to their 
value as decimals. The sum will be the required natural logarithm. 

Example. Find the natural logarithm corresponding to the common logarithm 
.497149. 

COMMON LOGARITHMS NATURAL LOGARITHMS 



.49 
.0071 
.000049 
.497149 



1.1282667 
.016348354 
.00011282667 

1.14472788067 



COM- 
MON 
LOGA- 
RITHM 


NATURAL 
LOGARITHM 


COM- 
MON 
LOGA- 
RITHM 


NATURAL 
LOGARITHM 


COM- 
MON 
LOGA- 
RITHM 


NATURAL 
LOGARITHM 


COM- 
MON 
LOGA- 
RITHM 


NATURAL 
LOGARITHM 


1 


2.30259 


26 


59.86721 


51 


117.43184 


76 


174.99647 


2 


4.60517 


27 


62.16980 


52 


119.73442 


77 


177.29905 


3 


6.90776 


28 


64.47238 


53 


122.03701 


78 


179.60164 


4 


9.21034 


29 


66.77497 


54 


124.33959 


79 


181.90422 


5 


11.51293 


30 


69.07755 


55 


126.64218 


80 


184.20681 


6 


13.81551 


31 


71.38014 


56 


128.94477 


81 


186.50939 


7 


16.11810 


32 


73.68272 


57 


131.24735 


82 


188.81198 


8 


18.42068 


33 


75.98531 


58 


133.54994 


83 


191.11456 


9 


20.73327 


34 


78.28789 


59 


135.85252 


84 


193.41715 


10 


23.02585 


35 


80.59048 


60 


138.15511 


85 


195.71973 


11 


25.32844 


36 


82.89306 


61 


140.45769 


86 


198.02232 


12 


27.63102 


37 


85.19565 


62 


142.76028 


87 


200.32490 


13 


29.93361 


38 


87.49823 


63 


145.06286 


88 


202.62749 


14 


32.23619 


39 


89.80082 


64 


147.36545 


89 


204.93007 


15 


34.53878 


40 


92.10340 


65 


149.66803 


90 


207.23266 


16 


36.84136 


41 


94.40599 


66 


151.97062 


91 


209.53524 


17 


39.14395 


42 


96.70857 


67 


154.27320 


92 


211.83783 


18 


41.44653 


43 


99.01116 


68 


156.57579 


93 


214.14041 


19 


43.74912 


44 


101.31374 


69 


158.87837 


94 


216.44300 


20 


46.05170 


45 


103.61633 


70 


161.18096 


95 


218.74558 


21 


48.35429 


46 


105.91891 


71 


163.48354 


96 


221.04817 


22 


50.65687 


47 


108.22150 


72 


165.78613 


97 


223.35075 


23 


52.95946 


48 . 


110.52408 


73 


168.08871 


98 


225.65334 


24 


55.26204 


49 


112.82667 


74 


170.39130 


99 


227.95592 


25 


57.56463 


50 


115.12925 


75 


172.69388 


100 


230.25851 



TABLES 



xxxni 



TABLE X 
FUNCTIONS OF ANGLES 



ANGLE 


SIN 


TAX 


SEC 


COS EC 


COT 


Cos 







0. 


0. 


.0 


CO 


CO 


1. 


90 


1 


0.0175 


0.0175 


1.0001 


57.299 


57.290 


0.9998 


89 


2 


.0349 


.0349 


1.0006 


28.654 


28.636 


.9994 


88 . 


8 


.0523 


.0524 


1.0014 


19.107 


19.081 


.9986 


87 


4 


.0698 


.0699 


1.0024 


14.336 


14.301 


.9976 


86 


5 


.0872 


.0875 


1.0038 


11,474 


11.430 


.9962 


85 


6 


0.1045 


0.1051 


1.0055 


9.5668 


9.5144 


0.9945 


84 


7 


.1219 


.1228 


1.0075 


8.2055 


8.1443 


.9925 


83 


8 


.1392 


.1405 


1.0098 


7.1853 


7.1154 


.9903 


82 


9 


.1564 


.1584 


1.0125 


6.3925 


6.3138 


.9877 


81 


10 


.1736 


.1763 


1.0154 


5.7588 


5.6713 


.9848 


80 


11 


0.1908 


0.1944 


.0187 


5.2408 


5.1446 


0.9816 


79 


12 


.2079 


.2126 


.0223 


4.8097 


4.7046 


.9781 


78 


13 


.2250 


.2309 


.0263 


4.4454 


4.3315 


.9744 


77 


14 


.2419 


.2493 


.0306 


4.1336 


4.0108 


.9703 


76 


15 


.2588 


.2679 


.0353 


3.8637 


3.7321 


.9659 


75 


16 


0.2756 


0.2867 


1.0403 


3.6280 


3.4874 


0.9613 


74 


17 


.2924 


.3057 


1.0457 


3.4203 


3.2709 


.9563 


73 


18 


.3090 


.3249 


1.0515 


3.2361 


3.0777 


.9511 


72 


19 


.3256 


.3443 


1.0576 


3.0716 


2.9042 


.9455 


71 


20 


.3420 


.3640 


1.0642 


2.9238 


2.7475 


.9397 


70 


21 


0.3584 


0.3839 


.0712 


2.7904 


2.6051 


0.9336 


69 


22 


.3746 


.4040 


.0785 


2.6695 


2.4751 


.9272 


68 


23 


.3907 


.4245 


.0864 


2.5593 


2.3559 


.9205 


67 


24 


.4067 


.4452 


.0946 


2.4586 


2.2460 


.9135 


66 


25 


.4226 


.4663 


.1034 


2.3662 


2.1445 


.9063 


65 


26 


0.4384 


0.4877 


.1126 


2.2812 


2.0503 


0.8988 


64 


27 


.4540 


.5095 


.1223 


2.2027 


1.9626 


.8910 


63 


28 


.4695 


.5317 


.1326 


2.1301 


1.8807 


.8829 


62 


29 


.4848 


.5543 


.1434 


2.0627 


1.8040 


.8746 


61 


30 


.5000 


.5774 


.1547 


2.0000 


1.7321 


.8660 


60 


81 


0.5150 


0.6009 


.1666 


1.9416 


1 .6643 


0.8572 


59 


32 


.5299 


.6249 


.1792 


1.8871 


1.6003 


.8480 


58 


88 


.5446 


.6494 


1.1924 


1.8361 


1.5399 


.8387 


57 


34 


.5592 


.6745 


1.2062 


1.7883 


1.4826 


.8290 


56 


35 


.5736 


.7002 


1.2208 


1.7435 


1.4281 


.8192 


55 


36 


0.5878 


0.7265 


1.2361 


1.7013 


.3764 


0.8090 


54 


37 


.6018 


.7536 


1.2521 


1.6616 


.3270 


.7986 


53 


88 


.6157 


.7813 


1.2690 


1.6243 


.2799 


.7880 


52 


39 


.6293 


.8098 


1.2868 


1.5890 


.2349 


.7771 


51 


40 


.6428 


.8391 


1.3054 


1.5557 


.1918 


.7660 


50 


41 


0.6561 


0.8693 


1.3250 


1.5243 


.1504 


0.7547 


49 


42 


.6691 


.9004 


1.3456 


1.4945 


.1106 


.7431 


48 


43 


.6820 


.9325 


1.3673 


1.4663 


.0724 


.7314 


47 


44 


.6947 


.9657 


1.3902 


1.4396 


.0355 


.7193 


46 


45 


.7071 


1. 


1.4142 


1.4142 





.7071 


45 




Cos 


COT 


COSEC 


SEO 


TAN 


SIN 


ANGLE 



XXXIV 



STRENGTH OF MATERIALS 



TABLE XI 
BENDING MOMENT AND SHEAR DIAGRAMS 





48 El 



7? Pb R Pa 

B 1 = y . R 2 = . 

M a = M c = 0. 

M " = F T 



Pi 



MOMENT 



SHEAR 



ti- 



B, = E 2 = P. 
3f a = M c = 0. 
3f & = ME = P a . 



* j * 





384 El 




2 wl 
=R l d- P(d-a). 



TABLES 



XXXV 




MOMENT 5 



SHEAR 
2 Pd s . 3 PcZ 2 





SHEAR 




~-- J >i 




SHEAR 

= wl. 
-wP 



1 ^ 



=~^-^r- 4- 




SHEAB 



E = wl + P. 



STRENGTH OF MATERIALS 




STRENGTH OF MATERIALS 

PART I 
MECHANICS OP MATERIALS 



PART I 
MECHANICS OF MATERIALS 

CHAPTER I 

ELASTIC PROPERTIES OF MATERIALS 

1. Introductory. In mechanics all bodies considered are assumed 
to be perfectly rigid; that is to say, it is assumed that no matter 
what system of forces acts on a body, the distance between any two 
points of the body remains unchanged. 

It has been found by experiment, however, that the behavior of 
natural bodies does not verify this assumption. Thus experiment 
shows that when a body formed of any substance whatever is acted 
upon by external forces it changes its shape more or less, and that 
when this change of shape becomes sufficiently great the body breaks. 
It has also been found that the amount of change in shape necessary 
to cause rupture depends on the material of which the body is made. 
For instance, a piece of vulcanized rubber will stretch about eight 
times its own length before breaking, while if a piece of steel is 
stretched until it breaks, the elongation preceding rupture is only 
from y 1 ^ to ^ of its original length. 

2. Subject-matter of the strength of materials. Since the assump- 
tion of rigidity upon which mechanics is based cannot be extended 
to natural bodies, mathematical analysis alone is not sufficient to 
determine the strength of any given structure. A knowledge of the 
physical properties peculiar to the material of which the structure is 
made is also essential. 

The subject-matter of the strength of materials, therefore, consists 
of two parts. First, a mathematical theory of the relation between 
the external forces which act on a body and its resultant change of 
shape, by means of which the direction and intensity of the forces 
acting at any point of the body may be calculated ; and, second, an 

1 



2 &TKENGTH OF MATERIALS 

experimental determination of the physical properties, such as 
strength and elasticity, of the various materials used in construction. 
Although it is convenient to divide the subject in this way, it 
must be understood that the two parts are, in reality, inseparable ; for 
the mathematical discussion involves physical constants which can 
be found only by experiment, while, on the other hand, experiment 
alone is powerless to determine the form which should be given to 
construction members in order to secure efficiency of design with 
economy of material. 

3. Stress, strain, and deformation. Whenever an external force 
acts on a body it creates a resisting force within the body. This, in 
fact, is simply another way of stating Newton's third law of motion, 
that to every action there exists a reaction equal in magnitude and 
opposite in direction. This internal resistance is due to innumerable 
small forces of attraction exerted between the molecules of the body, 
called " molecular forces," or stresses. A body subjected to the action 
of stress is said to be strained, and the resulting change in shape is 
called the deformation. 

For example, suppose a copper wire 40 in. long supports a weight of 10 Ib. and 
is stretched by this weight so that its length becomes 40.1 in. Then the sum of 
the stresses acting on any cross section of the wire is 10 Ib. , and the effect of this 
stress is to strain the wire until its deformation, or increase in length, is . 1 in. 

4. Tension, compression, and shear. In order to determine the 
relation between the stresses at any point in a solid body, only a 
small portion of the body is considered at a time, say an infinitesimal 
cube. This small cube is then assumed to act like a rigid body, and 
the relations between the stresses which act on it are determined by 
means of the conditions of equilibrium deduced in mechanics. 

By the principle of the resolution of forces, the stresses acting on 
any face of such an elementary cube can be analyzed into two com- 
ponents, one perpendicular to the face of the cube and the other 
lying in the plane of the face. That component of the stress which 
is perpendicular to the face of the cube is called the normal stress. 
If the normal stress pulls on the cube, and thus tends to increase its 
dimensions, it is called tension; if it pushes on the cube, and thus 
tends to decrease its dimensions, it is called compression. Tension is 
indicated by the sign -f and compression by the sign . 



ELASTIC PKOPEKTIES OF MATERIALS 



That component of the stress which lies in the plane of the face 
tends to slide this face past the adjoining portion of the body, and 
for this reason is called the shear, since its action resembles that of a 
pair of scissors or shears. 

5. Unit stress. If the total stress acting on any cross section of 
a body is divided by the area of the cross section, the result is the 
stress per unit of area, or unit stress. In what follows p will be used 
to denote the unit normal stress and q to denote the unit shear. 
Thus if a bar 2 in. square is stretched by a force of 800 lb., the 
unit normal stress is 

800 lb. 



4 in. 5 



= +2001b./in. 2 * 



If a rod is subjected to tension, it is customary to assume that the 
stress is uniformly distributed over any cross section of the rod. 
This assumption, however, is only approximately correct ; for if two 
parallel lines are drawn near the center of a 
rubber test piece, as ab and cd in Fig. 1, A, it 
is found that when the test piece is subjected 
to tension these two lines become convex 
toward one another, as indicated in Fig. 1, B, 
showing that the tensile stress is greater near 
the edges of the piece than at the center. In 
such a case of nonuniform distribution of 
stress, the smaller the area considered the 
nearer the unit stress approaches its true 
value. That is to say, if AP is the stress acting on a small area 

, then, in the notation of the calculus, 

AP dP 




FIG. 1 



Problem 1. A post 1 ft. in diameter supports a load of one ton.f Assuming 
that the stress is uniformly distributed over any cross section, find the unit 
normal stress. 

Problem 2. A shearing force of 50 lb. is uniformly distributed over an area 
4 in. square. Find the unit shear. 

* For the sake of brevity and clearness all dimensions in this book will be expressed 
as above ; that is, " lb. per sq. in." will be written " lb. /in.' 2 ," etc. 

t Throughout this book the word " ton " is used to denote the net ton of 2000 lb. 



4 STKENGTH OF MATEEIALS 

6. Unit deformation. If a bar of length I is subjected to tension 
or compression, its length is increased or diminished by a certain 
amount, say A. The ratio of this change in length to the original 
length of the bar is called the unit deformation, and will be denoted 
by s. Thus A ^ 

S = T' 

In other words, the unit deformation is the elongation or contraction 
per unit of length, or the percentage of deformation, and s is there- 
fore an abstract number. 

Problem 3. A copper wire 100 ft. long and .025 in. in diameter stretches 2.16 in. 
when pulled by a force of 15 Ib. Find the unit elongation. 

Problem 4. If the wire in Problem 3 was 250 ft. long, how much would it 
lengthen under the same pull ? 

Problem 5. A vertical wooden post 30 ft. long and 8 in. square shortens 
.00374 in. under a load of half a ton. What is its unit contraction? 

7. Strain diagrams. As mentioned in Article 1, experiment has 
shown that the effect of the action of external forces upon a body is 
to produce a change in its shape. If the body returns to its original 
shape when these external forces are removed it is said to be elastic, 
whereas if it remains deformed it is said to be plastic. 

For instance, the steel hairspring of a watch is an example of an 
elastic body, for although it is compressed thousands of times daily 
it returns each time to its original shape when the compressive force 
is removed. Wood, iron, glass, and ivory are other examples of elastic 
substances. 

As examples of plastic bodies may be taken such substances as 
putty, lead, and wet clay, for such materials retain any shape into 
which they may be pressed. 

It has been found by experiment that most of the materials used 
in engineering are almost perfectly elastic, if the forces acting on 
them are not too large. That is to say, if the external forces do not 
surpass a certain limit, the permanent deformation, although not 
zero, is so small as to be negligible. If, however, the external forces 
gradually increase, there comes a time when the body no longer 
regains its original form completely upon removal of the stress, but 
takes a permanent "set" due to plastic deformation. If the exter- 
nal forces increase beyond this point, the permanent (or plastic) 



ELASTIC PROPERTIES OF MATERIALS 5 

deformation also increases ; or, in other words, the tendency of the 
body to return to its original form grows less and less until rupture 
occurs. 

For example, suppose that a rod of steel or wrought iron is 
stretched by a tensile force applied at its ends. Then if the unit 
tensile forces acting on the rod are plotted as ordinates and the cor- 
responding unit elongations of the rod as abscissas, a curve will be 
obtained, as shown in Fig. 2.* 

Consider the curve for wrought iron obtained in this way. For 
stresses less than a certain amount, indicated by the ordinate at A in 



00 
































/ 


HARD 


STEE 


', 


















/ 


' 




















^ 




/ 






















30 

B 


j- 




. 1 


^-~ 





* 


MEDI 


JM STEEL 
n 


i 






| 




/* 


- 


~- - 








WROI 


GHT IRON 


* 


* 


^Z> 


^ |2 


^ 


^ 






















| ^ 


^j " 7} 

/'CAS 


T IROI 


r 




















K 






















































2 4 6 8 10 12 14 16 18 20 22 24 

EXTENSION, PER CENT 

FIG. 2 

Fig. 2, the deformation is very slight and is proportional to the stress 
which produces it, so that this portion OA of the strain diagram is a 
straight line. For stresses above A the deformation increases more 
rapidly than the stress which produces it, and consequently the strain 
diagram becomes curved. When the stress reaches a certain point B 
the material suddenly yields, the deformation increasing to a marked 
extent without any increase in the stress. Beyond this point the 
deformation increases with growing rapidity until rupture is aJoout 

* Drawn from data given in the United States Government Reports on Tests of Metals. 



6 STRENGTH OF MATERIALS 

to take place. At this stage of the experiment, indicated by C on the 
diagram, the material in the neighborhood of the place where rupture 
is to occur begins to draw out very rapidly, and in consequence the 
cross section of the piece diminishes at this point until rupture occurs. 

Within the portion OA of the strain diagram the stress is pro- 
portional to the deformation produced, and the material may be con- 
sidered to be perfectly elastic. For this reason the point A, which is 
the limit of proportionality of stress to deformation, is called the 
elastic limit The point B, at which the first signs of weakening occur, 
is called the yield point 

In commercial testing the tests are usually conducted so hurriedly 
that the position of the point A is not noted, and consequently the 
yield point is often called the elastic limit. The yield point, however, 
is not the true elastic limit, because plastic deformation begins to be 
manifested before this point is reached, namely, as soon as the stress 
passes A. 

At C the tangent to the strain curve is horizontal. Therefore the 
ordinate at this point indicates the maximum stress preceding rup- 
ture, which is called the ultimate strength of the material. 

8. Hooke's law and Young's modulus. The fact that within the 
elastic limit the deformation of a body is proportional to the stress 
producing it was discovered in 1678 by Robert Hooke, and is there- 
fore known as Hooke' s law. It can be stated by saying that the ratio 
of the unit stress to the unit deformation is a constant ; or, expressed 

as a formula, 

P _ 
7 = E > 

where E is a constant called the modulus of elasticity. E is also called 
Young's modulus, from the name of the first scientist who made any 
practical use of it. 

Since s is an abstract number, E has the same dimensions as p 
and is therefore expressed in lb./in. 2 Geometrically E is the slope 
of the line OA in Fig. 2, 

The answers given to the following problems were obtained by 
using the average values of Young's modulus given in Table I. 

Problem 6. A steel cable 500 ft. long and 1 in. in diameter is pulled by a force 
of 26 tons. How much does it stretch, and what is its unit elongation ? 



ELASTIC PROPERTIES OF MATERIALS 7 

Problem 7. A copper wire 10 ft. long and .04 in. in diameter is tested and found 
to stretch .289 in. under a pull of 50 Ib. What is the value of Young's modulus 
for copper deduced from this experiment ? 

Problem 8. A round cast-iron pillar 18 ft. high and 10 in. in diameter sup- 
ports a load of 12 tons. How much does it shorten, and what is its unit con- 
traction ? 

Problem 9. A wrought-iron bar 20 ft. long and 1 in. square is stretched .266 in. 
What is the force acting on it ? 

9. Poisson's ratio. It has been found by experiment that when 
a rod is subjected to tension or compression its transverse dimensions 
are changed as well as its length. For instance, if a round rod is in 
tension, it increases in length and decreases in diameter, whereas, if 
the rod is compressed, it decreases in length and increases in diam- 
eter. Experiment has also shown that this lateral contraction or 
expansion is proportional to the change in length of the bar; that 
is to say, the ratio of the unit lateral deformation to the unit change 

in length is constant, say This constant is called Poisson's ratio. 

m 

from the name of its originator. 

Poisson's ratio varies somewhat for different materials, but ordi- 
narily lies between J- and |. Values of this ratio for a number of 
materials are given in Table I. 

Problem 10. What is the lateral contraction of the bar in Problem 9 ? 
Problem 11. A soft steel cylinder 1 ft. high and 2 in. in diameter bears a 
weight of 75 tons. How much is its diameter increased ? 

10. Ultimate strength. From the definition given in Article 7, 
the ultimate strength of a body is the greatest unit stress it can stand 
without breaking. In calculating the ultimate strength no account 
is taken of the lateral contraction or expansion of the body, the ulti- 
mate strength being denned as the breaking load divided by the 
original area of a cross section of the piece before strain. The reason 
for this arbitrary definition of the ultimate strength is that the actual 
load on any member of an engineering structure usually lies within 
the elastic limit of the material, and within this limit the change 
in area of a cross section of the member is so small that it can be 
neglected. 

Tabulated values of the ultimate strength of various materials in 
tension, compression, and shear are given in Table I. 



8 STEENGTH OF MATEEIALS 

Problem 12. How great a pull can a copper wire .2 in. in diameter stand with- 
out breaking ? 

Problem 13. How large must a square wrought-iron bar be made to stand a 
pull of 3000 lb.? 

Problem 14. A mild steel plate is J in. thick. How wide must it be to stand a 
pull of 1 ton ? 

Problem 15. A round wooden post is 6 in. in diameter. How great a load will 
it bear ? 

11. Elastic law. Certain substances, notably cast iron, stone, 
cement, and concrete, do not conform to Hooke's law, in that the 
deformation is not proportional to the stress which produces it. 
Consequently, for such substances the strain diagram is nowhere a 
straight line, but is curved throughout, as shown in the curve for 
cast iron in Fig. 2. In this case the modulus of elasticity changes 
from point to point. 

In the reports of the U. S. Testing Laboratory at the Watertown 
Arsenal, the modulus of elasticity is defined as the quotient of the 
unit stress by the unit deformation minus the permanent set. Thus, 

if s r denotes the permanent set, this definition makes E = *- - 

s s 

Numerous attempts have been made to determine the equation 
of the strain curve for various materials which do not conform to 
Hooke's law, and a corresponding number of formulas, or elastic laws, 
have been proposed. The one which agrees best with experiment is 
the exponential law, expressed by the formula 

s = vp*, 

where v and cr are constants determined by experiment. From Bach's 
experiments the values of v and cr were found to be such that 

for cast iron in tension, s = 24)26 1 8>000 P l ' 6 6 3 5 

for cast iron in compression, s = 18>46 1 9>200 P 1 ' 3 9 6 > 
the unit stress p being expressed in lb./in. 2 

However, all such elastic laws are at best merely interpolation 
formulas which are approximately true within the limits of the 
experiments from which they were obtained. For this reason it is 
best to carry out all investigations in the strength of materials 
under the assumption of Hooke's law, and then modify the results 
by a factor of safety, as explained in Article 21. 



ELASTIC PEOPEETIES OF MATERIALS 9 

12. Classification of materials. Materials ordinarily used in engi- 
neering construction may be divided into three classes, plastic, 
supple, and elastic. 

Plastic materials are characterized by their inability to resist stress 
without receiving permanent deformation. Examples of such mate- 
rials are lead, wet clay, mortar before setting, etc. 

Supple bodies are characterized by their lack of stiffness. In other 
words, supple bodies are capable of undergoing large amounts of 
elastic deformation without receiving any plastic deformation. In 
this respect plastic and supple bodies exhibit the two extremes of 
physical behavior. Examples of supple bodies are rubber, copper, 
rope, cables, textile fabrics, etc. 

Elastic bodies comprise all the hard and rigid substances, such 
as iron, steel, wood, glass, stone, etc. For such bodies the plastic 
deformation for any stress within the elastic limit is so small as 
to be negligible ; but when the stress surpasses this limit the plastic 
deformation becomes measurable and gradually increases until rup- 
ture occurs. This permanent deformation is the outward manifes- 
tation of a change in the molecular arrangement of the body. For 
a stress within the elastic limit the forces of attraction between the 
molecules are sufficiently great to hold the molecules in equilibrium ; 
but when the stress surpasses the elastic limit, the molecular forces 
can no longer maintain equilibrium and a change in the relation 
between the molecules of the body takes place, which results in the 
body taking a permanent set. 

Eigid bodies have the character of supple bodies when one of 
their dimensions is very small as compared with the others. An 
instance of this is the flexibility of an iron or steel wire whose 
length is very great as compared with its diameter. Furthermore, 
rigid bodies behave like plastic bodies when their temperature is 
raised to a certain point. For example, when iron and steel are 
heated to a cherry redness- they become plastic and acquire the 
property of uniting by contact. 

13. Time effect. It has been found by experiment that elastic 
deformation is manifested simultaneously with the application of a 
stress, but that plastic deformation does not appear until much later. 
Thus if a constant load acts for a considerable time, the deformation 



10 STRENGTH OF MATERIALS 

gradually increases; and when the load is removed the return of 
the body to its original configuration is also gradual. This phenom- 
enon of the deformation lagging behind the stress which produces 
it is called hysteresis. The gradual increase in the deformation 
under constant stress is also called the flow of the material; and 
the gradual return of the body to its original shape upon removal 
of the stress is known as elastic afterwork. This gradual flow which 
occurs under constant stress approaches a limit if the stress lies 
below the elastic limit, but continues up to fracture if the stress is 
sufficiently great. 

14. Fatigue of metals. If a stress lies well within the elastic limit, 
it can be removed and repeated as often as desired without causing 
rupture. If, however, a metal is stressed beyond the elastic limit, 
and this stress is removed and repeated, or alternates between tension 
and compression, a sufficient number of times, it will eventually cause 
rupture. This phenomenon is known as the fatigue of metals, and has 
been made the subject of laborious experiment by Wbhler, Bauschin- 
ger, and others. The results of their experiments show that the less 
the range of variation of stress, the greater the number of repetitions 
or reversals of stress necessary to produce rupture. Among other 
results Bauschinger found that for cast iron with an ultimate ten- 
sile strength of 64,100 lb./in. 2 , the maximum tensile stress which 
could be removed and repeated indefinitely without causing rupture 
was 35,300 lb./in. 2 ; and that the maximum stress which could be 
alternated indefinitely between tension and compression of equal 
amounts without causing rupture was 29,100 lb./in. 2 For other 
kinds of iron and steel Bauschinger obtained similar results, the 
limit of reversible stress in each case agreeing closely with the elastic 
limit. From this we conclude that the elastic limit of a material is 
much more important than its ultimate strength in determining the 
stability of an engineering structure of which it forms a part. 

The fatigue of metals indicates that dislocation of matter begins 
to be produced as soon as the elastic limit is passed, and continues 
under the action of relatively small forces. This is confirmed by the 
well-known fact that if, as the result of a blow, a fissure or crack is 
started in a piece of glass or cast iron, this fissure will spread with- 
out any apparent cause until the piece breaks in two, the only way 



ELASTIC PROPERTIES OF MATERIALS H 

of stopping this tendency to spread being by boring a small hole at 
either end of the fissure. 

The explanation of the above is that for stresses within the elastic 
limit the temperature of the body is not raised, and consequently all 
the work of deformation is stored up in the body to be given out 
again in the form of mechanical energy upon removal of the stress. 
If, however, the elastic limit is surpassed, the friction of the mole- 
cules sliding on each other generates a certain amount of heat, and 
the energy thus transformed into heat is not available for restoring 
the body to its original configuration. 

15. Hardening effects of overstraining. When such materials as 
iron and steel are stressed beyond the elastic limit, it is found upon 
removal of the stress that the effect of this overstrain is a hardening 
of the material, and that this hardening increases indefinitely with 
time. For example, if a plate of soft steel is cold punched, the 
material surrounding the hole is severely strained. After an interval 
of rest the effects of this overstrain is manifested in a hardening of 
the material which continues to increase for months. If the plate is 
subsequently stressed, the inability of the portion .overstrained to 
yield with the rest of the plate causes the stress to be concentrated 
on these portions, and results in a serious weakening of the plate. 

Other practical instances of hardening due to overstrain are found 
in plates subjected to shearing and planing, armor plates pierced by 
cannon balls, plates and bars rolled, hammered, or bent when cold, 
wire cold drawn, etc. 

16. Fragility. In the solidification of melted bodies different 
parts are unequally contracted or expanded. This gives rise to in- 
ternal stresses, or what is called latent molecular action, and puts the 
body in a state of strain without the application of any external 
forces. For instance, if a drop of melted glass is allowed to fall into 
water, the outside of the drop is instantly cooled and consequently 
contracted, while the inside still remains molten. Since the part 
within cannot contract while molten, the contraction of the outside 
causes such large internal stresses that the glass is shattered. 

Bodies in which latent molecular action exists have the character 
of an explosive, in that they are capable of standing a large static 
stress but are easily broken by a blow, and for this reason they are 



12 STEENGTH OF MATEEIALS 

called brittle or fragile. The explanation of fragility is that the vibra- 
tions caused by a blow are reinforced by the latent internal stresses 
until rupture ensues. 

17. Initial internal stress. In certain bodies, such as cast iron, 
stone, and cement, a state of internal stress may exist without the 
application of any external force. This initial internal stress may be 
the result of deformation caused by previously applied loads, or may 
be occasioned by temperature changes, as mentioned in the preceding 
article. The first load applied to such bodies gives them a slight 
permanent deformation, but under subsequent loads their behavior 
is completely elastic. The first load, in this case, serves to relieve 
the strain due to initial internal stress, and consequently the behavior 
of the body under subsequent loads is normal. A body which is free 
from internal stress is said to be in a " state of ease," a term which 
is due to Professor Karl Pearson. 

18. Annealing. The process of annealing metals consists in heat- 
ing them to a cherry redness and then allowing them to cool slowly. 
The effect of this process is to relieve any initial internal stress, or 
stress due to overstrain, and put the material in a state of ease. 
Hardening due to overstrain is of frequent occurrence in engineering, 
and the only certain remedy for it is annealing. If this is imprac- 
ticable, hardening can be practically avoided by substituting boring 
for punching, sawing for shearing, etc. 

19. Temperature stresses. A property especially characteristic of 
metals is that of expansion with rise of temperature. The proportion 
of its length which a bar expands when its temperature is raised one 
degree is called the coefficient of linear expansion, and will be denoted 
by L. The following table gives the value of L for one degree 
Fahrenheit for the substances named. 

Steel, hard L = .0000074 

u soft L = .0000061 

Iron, cast L = .0000063 

(t wrought L = .0000068 

Timber L = .0000028 

Granite L = .0000047 

Sandstone L = .0000065 

If a body is fixed to immovable supports so that when the temper- 
ature of the body is raised these supports prevent it from expanding, 



ELASTIC PKOPERTIES OF MATERIALS 13 

stresses are produced in the body called temperature stresses. Thus 
suppose a bar of length / is rigidly fastened to immovable supports 
and its temperature is then raised a certain amount. Let A/ be the 
amount the bar would naturally lengthen under this rise in temper- 
ature if left free to move. Then the stress necessary to produce a 
shortening of this amount is the temperature stress. 
If the temperature of the bar is raised T degrees, 

A/ = LIT, 
and consequently s = = LT. 

L 

Therefore, if p denotes the unit temperature stress, 
p = sE = LTE. 

The temperature of metals also has a marked influence upon 
their ultimate strength. Experiments along this line show that at 
296 F. the tensile strengths of iron and steel are about twice as 
great as at ordinary temperatures. 

Problem 16. A wrought-iron bar is 20 ft. long at 32 F. How long will it be at 
95 F. ? 

Problem 17. A cast-iron pipe 10 ft. long is placed between two heavy walls. 
What will be the stress in the pipe if the temperature rises 25 ? 

Problem 18. Steel railroad rails, each 30 ft. long, are laid at a temperature of 
40 F. What space must be left between them in order that their ends shall just 
meet at 100 F. ? 

Problem 19. In the preceding problem, if the rails are laid with their ends in 
contact, what will be the temperature stress in them at 100 F. ? 

20. Effect of length, diameter, and form of cross section. When 
an external force is first applied to a body the internal stress is dis- 
tributed uniformly throughout the body and, consequently, all parts 
are equally deformed.* When the stress surpasses the elastic limit 
this is no longer true, and certain portions of the body begin to mani- 
fest greater deformation than others. For instance, consider a bar of 
soft steel under tension. As the stress increases from zero to the 
elastic limit the bar gradually lengthens and its cross section dimin- 
ishes, all parts being equally affected. When the stress passes beyond 
the elastic limit the cross section at some particular point of the bar, 

* This depends somewhat upon the way the external force is applied. 



14 



STRENGTH OF MATERIALS 




usually near the center, begins to diminish more rapidly than else- 
where. This contraction of section intensifies the unit stress at this 
point, and this in turn tends to a still greater reduction of section until 

finally rupture occurs. 

The appearance of a 
bar subjected to a test of 
this kind is represented in 
Fig. 3. The contracted por- 
tion, AB, of the bar is called the region of striction. The contraction 
of the section at which rupture occurs is usually considerable ; for 
soft steel its amount is from .4 to .6 of the original area of the bar. 

In Article 6 the unit elongation was defined as the ratio of the 
total elongation to the original length of the bar. It has been found 
by experiment, however, that the extent of the region of striction 
depends on the transverse dimensions of the bar and not on its length, 
the region of striction increasing in extent as the transverse dimen- 
sions of the bar increase. Consequently, if two bars are of equivalent 
cross section but of different lengths, the region of striction will be 
the same for both, and therefore the unit elongation will appear to 
be less for the long bar than for the short one. On the other hand, 
if the two bars are of the same length, but one is thicker than the 
other, the region of striction will be longer for the thick bar, and 
therefore the unit elon- 



gation of this bar will 
appear to be greater 
than for the other. 

The form of cross sec- 
tion of test pieces sub- 
jected to tensile tests 
has also an important 
influence on their elon- 
gation and on their ulti- 
mate strength. If a sharp change in cross section occurs at any 
point, nonductile materials, such as cast iron, will break at this sec- 
tion under a smaller unit stress than they could otherwise carry. 
This is due to a greater intensity of stress at the section where the 
change in area occurs. 




FIG. 4 



ELASTIC PROPERTIES OF MATERIALS 15 

For ductile materials, such as wrought iron and mild steel, the 
striction extends over a length six or eight times the width of the 
piece. Consequently, if the test piece has a form similar to one 
of those represented in Fig. 4, in which the length AB is less than 
six or eight times the width of the piece, the flow of the metal is 
restrained and therefore its ultimate strength is raised. This has an 
important bearing on the strength of riv- 
eted plates subjected to tensile strain. It f | 

has been experimentally proved that such 
plates will stand a greater tension than 
plates of uniform cross section whose 
sectional area is equal to the sum of the 
sectional areas between the rivet holes. 

In Article 10 the ultimate strength 

i 

was denned as the ratio of the maximum 

FIG. 5 

stress to the original sectional area of 

the bar. It is evident from what precedes, therefore, that the unit 
elongation and the ultimate strength are not absolute quantities, but 
depend on the form of the test piece and the conditions of the test. 
For this reason it is absolutely essential that the results of any test 
be accompanied by an accurate description of the circumstances under 
which they were obtained. The elastic limit and modulus of elasticity, 
on the contrary, have an intrinsic value independent of their method 
of determination, and therefore more accurately define the elastic 
properties of any material. 

The tensile strength of long rods is affected in a way different from 
any of the preceding. Since no material is perfectly homogeneous, the 
longer the rod the greater the chance that a flaw will occur in it some- 
where. If, then, by numerous tests of short pieces, it has been deter- 
mined how much a material lacks of being homogeneous, the strength 
of a rod of this material of any given length can be calculated by 
means of the theory of probabilities. Such a theory has been worked 
out by Professor Chaplin* and verified experimentally. 

If one dimension of a body is very small compared with the 
others, as, for example, in long wires or very thin plates, the body 

*Van Nostrand's Eng. Mag., December, 1880; also Proc. Eng. Club, Philadelphia, 

March, 1882. 



16 STRENGTH OF MATERIALS 

may be permanently deformed by stresses below the elastic limit. 
The reason for this is that the smallest dimension of such a body 
is of the same order of magnitude as the deformation of one of the 
other dimensions, and consequently Hooke's law does not apply in 
this case. 

21. Factor of safety. In order to assure absolute stability to any 
structure it is clear from what precedes that the actual stresses 
occurring in the structure must not exceed the elastic limit of the 
material used. 

For many materials, however, it is very difficult to determine the 
elastic limit, while for other materials for which the determination 
is easier, such as iron and steel, the elastic limit is subject to large 
variations in value, and it is impossible to do more than assign wide 
limits within which it may be expected to lie. For this reason it is 
customary to judge the quality of a material by its ultimate strength 
instead of by its elastic limit, and assume a certain fraction of the 
ultimate strength as the allowable working stress. 

The number which expresses the ratio of the ultimate strength to 
the working stress is called the factor of safety. Thus 

ultimate strength 



Factor of safety = 



working- stress 



No general and rational method of determining the factor of safety 
can be given. For, in the first place, formulas deduced from theoret- 
ical considerations rest on the assumption that the material considered 
is perfectly elastic, homogeneous, and isotropic, an assumption which 
is never completely fulfilled. Such formulas give, therefore, only an 
approximate idea of the state of stress within the body. 

Moreover, the forms of construction members assumed for pur- 
poses of calculation do not exactly correspond to those actually used ; 
also certain conditions are unforeseen, and therefore unprovided for, 
such as the sinking of foundations, accidental shocks, etc. 

In metal constructions rust is another element which tends to 
reduce their strength, and in timber constructions the same is 
true of wet and dry rot. Care is usually taken to prevent rust and 
decay, but the preservative processes used never perfectly accomplish 
their object. 



ELASTIC PKOPEETIES OF MATERIALS 17 

Besides these elements of uncertainty every construction is 
attended by its own peculiar circumstances, such as the duration 
to be given to it, the gravity of an accident, etc., which requires a 
special determination of the factor of safety. 

For all these reasons it is impossible to definitely fix a factor of 
safety which will fit all cases, and the only guide that can be given 
as to its choice is to say that it will lie between certain limits. 
According to Resal,* the factor of safety for iron, steel, and ductile 
metals should be 4 or 3, and never less than 2J- ; for heterogeneous 
materials, such as cast iron, wood, and stone, the factor of safety 
should lie between 20 and 10, and never be less than the latter. 

Problem 20. In the United States government tests of rifle-barrel steel it was 
found that for a certain sample the unit tensile stress at the elastic limit was 71,000 
lb./in. 2 , and that the ultimate tensile strength was 118,000 lb./in. 2 What must 
the factor of safety be in order to bring the working stress within the elastic 
limit ? 

Problem 21. In the United States government tests of concrete cubes made 
of Atlas cement in the proportions of 1 part of sand to 3 of cement and 6 of broken 
stone, the ultimate compressive strength of one specimen was 883 lb./in. 2 , and of 
another specimen was 3256 lb./in. 2 If the working stress is determined from the 
ultimate strength of the first specimen by using a factor of safety of 5, what factor of 
safety must be used to determine the same working stress from the other specimen ? 

Problem 22. An elevator cab weighs 3 tons. With a factor of safety of 5, how 
large must a steel cable be to support the cab ? (Use Roebling's tables for wire 
rope given in Part II.) 

22. Work done in producing strain. In constructing the strain 
diagram, explained in Article 7, the unit stresses were plotted as 
ordinates and the corresponding unit deformations as abscissas. The 
autographic apparatus on a testing machine also gives a diagram 
which represents the strain, but in which the loads are the ordinates 
and the corresponding total deformations are the abscissas. The two 
diagrams are similar up to the elastic limit but not beyond this point, 
for after the elastic limit is passed, the area of cross section begins to 
change appreciably so that the unit stress is no longer proportional 
to the load. If, however, the unit stress is obtained by dividing the 
load by the original area of cross section, without taking into account 
the lateral deformation, the plotted strain diagram will be similar to 
the autographic load-deformation diagram. 

* Resal, Resistance des Mattriaux, p. 195. 



18 STRENGTH OF MATERIALS 

The load-deformation diagram has a special physical significance, 
namely that the area under the curve up to any point represents the 
work done in producing the strain up to that point. In this respect 
the autographic strain diagram resembles the indicator diagram on a 
steam engine. 

Since the elastic limit marks the limit within which the material 
may be considered as perfectly elastic, the area under the strain curve 
up to the elastic limit represents the amount of work which can be 
stored up in the form of potential energy, and is called the resilience 
of the test piece. Thus, if p denotes the unit stress at the elastic limit 
and F the area of cross section, the load is Fp ; and hence if A/ denotes 
the total deformation at the elastic limit, the work done up to this 

pi 
point is }fpF&l. From Hooke's law, A/ = Consequently the ex- 

1 p 2 lF 
pression for the resilience becomes > or, since IF represents the 

E 1 V *V 

volume V of the test piece, this may be written - The resil- 

1 p 2 2 E 

ience per unit volume, - > is called the modulus of elastic resilience 

of the material. 

EXERCISES ON CHAPTER I 

Problem 23. A f-in. wrought-iron bolt failed in the testing machine under a pull 

of 20,000 Ib. Diameter at root of thread = .5039 in. ; find its ultimate tensile strength. 

Problem 24. Four ^-in. steel cables are used with a block and tackle on the hoist 

of a crane whose capacity is rated at 6000 Ib. What is the factor of safety ? (Use 

Roebling's tables, Part II, for ultimate strength of rope.) 

Problem 25. A vertical hydraulic press weighing 100 tons is supported by four 
24-in. round cold-rolled steel rods. Find the factor of safety. 

Problem 26. A block and tackle consists of six strands of flexible ^-in. steel cable. 
What load can be supported with a factor of safety of 5 ? 

Problem 27. A wooden bar 6 ft. long, suspended vertically, is found to lengthen 
.013 in. under a load of 2100 Ib. hung at the end. Find the value of E for this bar. 

Problem 28. A copper wire in. in di- 
ameter and 500 ft. long is used as a crane 
trolley. The wire is stretched with a force 
of 100 Ib. when the temperature is 80 F. 
Find the pull in the wire when the temper- 
ature is F., and the factor of safety. 
U 7^ Problem 29. An extended shank is 

made for a ||-in. drill by boring a ||-in. 

hole in the end of a 10-in. length of cold-rolled steel, fitting the shank into this and 
putting a steel taper pin through both (Fig. 6). Standard pins taper \ in. per foot. 



n 



ELASTIC PROPERTIES OF MATERIALS 



19 



What size pin should be used in order that the strength of the pin against shear 
may equal the strength of the drill shank in compression around the hole ? 

Problem 30. The head of a steam cylinder of 12-in. inside diameter is held on by 
10 wrought-iron bolts. How tight should these bolts be screwed up in order that 
the cylinder may be steam tight under a pressure of 180 lb./in. 2 ? 

Problem 31. Find the depth of head of a wrought-iron bolt in terms of its 
diameter in order that the tensile strength of the bolt may equal the shearing 
strength of the head. 

Problem 32. The pendulum rod of a regulator 
used in an astronomical observatory is made of 
nickel steel in the proportion of 35.7 per cent nickel 
to 64.3 per cent steel. The coefficient of expansion 
of this alloy is approximately j 1 ^ that of steel, ^ 
that of copper, and ^ that of aluminum. This is 
tempered for several weeks, starting at 180 F. and 
gradually lowering to the temperature of the room, 
which eliminates the effect of elastic af terwork. 

The rod carries two compensation tubes, A 
and #, Fig. 7, one of copper and the other of 
steel, the length of the two together being 10 cm. 
If the length of the rod is 1 m., find the lengths of 
the two compensation tubes so that a change in 
temperature shall not affect the length of the 
pendulum. 

Problem 33. Refer to the Watertown Arsenal 
Reports ( United States Government Reports on Tests 
of Metals), and from the experimental results there tabulated draw typical strain 
diagrams for mild steel, wrought iron, cast iron, and timber, and compute E in 
each case. 

Problem 34. A steel wire \ in. in diameter and a brass wire in. in diameter 
jointly support a load of 1200 Ib. If the wires were of the same length when the 
load was applied, find the proportion of the load carried by each. 

Problem 35. An engine cylinder is 10 in. inside diameter and carries a steam 
pressure of 80 lb./in. 2 Find the number and size of the bolts required for the 
cylinder head for a working stress in the bolts of 2000 lb./in. 2 

Problem 36. Find the required diameter for a short piston rod of hard steel 
for a piston 20 in. in diameter and steam pressure of 125 lb./in. 2 Use factor of 
safety of 8. 

Problem 37. A rivet \ in. in diameter connects two wrought-iron plates each f 
in. thick. Compare the shearing strength of the rivet with the crushing strength of 
the plates around the rivet hole. 




FIG. 7 



CHAPTER II 



FUNDAMENTAL RELATIONS BETWEEN STRESS AND 
DEFORMATION 

23. Relations between the stress components. In order to deter- 
mine the relation between the stresses and deformations within an 
elastic body, it is necessary to make certain assumptions as to the 
nature of the body and the manner in which the external forces are 
applied to it. 

The first assumption to be made is that the material of which the 
body is composed is homogeneous ; that is to say, that the elastic 
properties of any two samples taken from different parts of the body 

are exactly alike. If, more- 
over, the surface of the body 
is continuous and the exter- 
nal forces are distributed con- 
tinuously over this surface, 
or, in other words, if there 
are no cracks or other sud- 
den changes of section in the 
body, and the external forces 
are distributed over a consid- 
erable bearing surface, it fol- 
lows, in consequence of the 

above assumptions, that the deformation at any point of the body is 
a continuous function of the coordinates of that point. In other 
words, under the above assumptions the deformation at any point 
of the body differs only infiniteshnally from the deformation at a 
neighboring point. 

Since, by Hooke's law, the stress is proportional to the deforma- 
tion, it follows that the stress is also distributed continuously 
throughout the body, that is, that the stress at any point of the 

' 20 




X. 



RELATIONS BETWEEN STKESS AND DEFOKMATION 21 



FIG. 9 



body differs only infinitesiinally from the stress at a neighboring 
point. This is called the law of continuity. 

Now consider an infinitesimal cube cut out of an elastic body 
which is subject to the above assumptions, and let the coordinate 
axes be taken along three adja- 
cent edges of the cube, as shown 
in Fig. 8. Then, from the law of 
continuity, the resultant of the 
normal stresses acting on any face 
of this cube is equal to their sum 
and is applied at the center of 
gravity of the face. Consequently, 
these resultants must all lie in one 
or other of the three diametral 
planes drawn through the center 
of the cube parallel to the coordi- 
nate planes. The stresses lying in 
any one of these planes, say the diametrical plane parallel to ZOX, 
will then be as represented in Fig. 9. 

Since the resultant normal stresses on opposite faces of the cube 
approach equality as the faces of the cube approach coincidence, we 
may write 

= and ='. 



For equilibrium against rotation 
the four shearing stresses must 
also be of equal intensity, and 
therefore 



By considering the other two 
diametral planes similar relations 
between the normal and shearing 
FIG. 10 stresses can be established. Con- 

sequently, the shearing stresses at 

any point in an elastic "body in planes mutually at right angles are 
of equal intensity in each of these planes. 




22 



STRENGTH OF MATERIALS 



24. Planar strain. If no stress occurs on one pair of opposite faces 
of the cube, the resultant stresses on the other faces all lie in one of 
the diametral planes. This is called the planar condition of strain. 

Suppose the -axis is drawn in the direction in which no stress 
occurs, as shown in Fig. 10. Then the stresses all lie in the plane 
parallel to XOY, and the relation between them is as represented in 
Fig. 9 of the preceding article. 

25. Stress in different directions. As an application of planar 
stress, consider a triangular prism on which no stress occurs in 
the direction of its length. Let the ^-axis be drawn in the direction 
in which no stress occurs, and let a denote the angle which the 




FIG. 11 



inclined face of the prism makes with the horizontal, as shown in 
Fig. 11. Then if dF denotes the area of the inclined face ABCD, 
and p', q f denote the normal and shearing stresses on this face 
respectively, p' and q 1 can be expressed in terms of p x) p y) and q by 
means of the conditions of equilibrium. Thus, from 2 hor. comps. = 0, 

p'dFsma + q'dFcosa p x dF sin a qdFcosa = 0. 

Similarly, from 2 vert, comps. = 0, 

p'dFcosa q'dFsma p y dF cos a qdFsma = 0. 

Dividing by dF, these equations become 

{ p 1 sma + q' cosa p x sma q cosa = 0, 
\p' cosa q f sma p y cosa q sma = 0. 



RELATIONS BETWEEN STRESS AND DEFORMATION 23 



Eliminating q', 

p' = p x sin 2 # + p v cos 2 a; + 2 q sin a; cos a;. 



From trigonometry, 
. 1 cos 2 a 



1 + cos 2 a 



, . 
2 sma cosa; = sin 2 a. 



Therefore, by substituting these values, 

(2) p f = Px ^ v + ^ y P* cos 2 a + q sin 2 a. 

2 2 

Similarly, by eliminating p r from equations (1), 
(3) 

Problem 38. At a certain point in a vertical cross section of a beam the unit 
normal stress is 300 lb./in. 2 , and the unit shear is 100 lb./in. 2 Find the normal 
stress and the shear at this point in a 
plane inclined at 30 to the horizontal. 

Solution. Suppose a small cube cut 
out of the beam at the point N (Fig. 
12). Then, by the theorem in Article 23, 
there will also be a unit shear of inten- 
sity q on the top and bottom faces of 
the cube. In the present case, there- 
fore, p x - 300 lb./in. 2 , p y = 0, and 
q = 100 lb./in. 2 Substituting these 
values in equations (2) and (3), and 
putting a = 30, the unit normal stress 
and unit shear on a plane through N inclined at 30 to the horizontal are 
p'= 161.5 lb./in. 2 , q' = 179.8 lb./in. 2 

26. Maximum normal stress. The condition that p 1 shall be a 

dr> ! 
maximum or a minimum is that -j- 0. Applying thie condition 

to equation (2), 




FIG. 12 



whence 

(5) 
and consequently 

(6) 



tan 2 a = 



2q 



a = - tan 




24 STRENGTH OF MATERIALS 

where X is zero or an arbitrary integer, either positive or negative. 
Equation (6) gives the angles which the planes containing the maxi- 
mum and minimum normal stresses make with the horizontal. 
From equation (5), 




Substituting these values of sin 2 a and cos 2 a in equation (2), the 
maximum and minimum values of the normal stress are found to be 





rain 

27. Principal stresses. Since X in equation (6) is an integer, the 
two values of a given by this equation differ by 90, and, conse- 
quently, the planes containing the maximum and minimum normal 
stresses are at right angles. The maximum and minimum normal 
stresses are called principal stresses, and the directions in which they 
act, principal directions. 

From equation (3), the right member of equation (4) is equal to 
2 q f . But since equation (4) is the condition for a maximum or min- 
imum value of the normal stress, it is evident that the normal stress 
is greatest or least when the shear is zero. 

The results of this article can therefore be summed up in the 
following theorem. 

Through each point of a body subjected to planar strain there are 
two principal directions at right angles, in each of which the shear is 
zero. 

Problem 39. Find the principal stresses and the principal directions at a point 
in a vertical cross section of a beam at which the unit normal stress is 400 lb./in. 2 
and the unit shear is 250 lb./in. 2 

Solution. In this problem p x = 400 lb./in. 2 , p v = 0, and q = 260 lb./in. 2 
Therefore, from equation (6), 

a = - tan-i - + ^ = - 25 40.2', or + 64 19.8'; 

and from equation (7), 

Pmax - 620 lb./in. 2 , p' min = - 120 lb./in. 2 



KELATIONS BETWEEN STEESS AND DEFORMATION 25 

28. Maximum shear. The condition that cf shall be a maximum or 
a minimum is that -~ = 0. Applying this condition to equation (3), 

CL OC 

= ^ y 2 cos 2 a 2 q sin 2 a ; 
whence 
(8) 

By comparing equations (5) and (8) it is evident that tan 2 a, 
from (8), equals cot 2 a, from (5). Therefore the values of 2 a 
obtained from these equations differ by 90, and hence the values 
of a differ by 45. Therefore the planes of maximum and mini- 
mum shear are inclined at 4^ t ^ e planes of maximum and 
minimum normal stress. 

From equation (8), 



Substituting these values of sin 2 a and cos 2 a in equation (3), 
the maximum and minimum values of the shear are found to be 



(9) ffJn = 



It is to be noticed that the maximum and minimum values of the 
shear given by equation (9) are equal in absolute amount and differ 
only in sign, which agrees with the theorem stated in Article 23. 

Problem 40. Find the maximum and minimum values of the shear in Prob- 
lem 39, and their directions. 

29. Linear strain. If a body is strained in only one direction, the 
strain is said to be linear. For instance, a vertical post supporting a 
weight, or a rod under tension, is subjected to linear strain. The 
unit normal stress and unit shear acting on any inclined section of 
a body strained in this way can be obtained by supposing the axes 
of coordinates drawn in the principal directions and putting q = 
and p y = in equations (2) and (3). These values can also be 
derived independently, as follows. 



26 



STRENGTH OF MATERIALS 



Consider an elementary triangular prism, and let the axis of X be 
drawn in the direction of the linear strain. The stresses acting on 

the prism will then be as shown in 
Fig. 13. Let dF denote the area of 
the inclined face. Then the area of 
the vertical face is dF sin a. Resolv- 
ing p x into components parallel to p' 
and q f respectively, the conditions of 
X- equilibrium are 

p x sin a (dF sin a) = p'dF, 
p x cos a (dF sin a) = q'dF; 




FIG. 13 



or, dividing by dF, 



(10) 



p : = 



?= 



dq' 



From the condition -- = 0, it is found that the maximum shear 
da 

occurs when a 45, and its value is 









For a or 90, q' = 0. Consequently, there is no shear in 
planes parallel or perpendicular to the direction of the linear strain. 

Problem 41. A wrought-iron bar 4 in. wide and fin. thick is subjected to a 
pull of 10 tons. What is the unit shear and unit normal stress on a plane inclined 
at 30 to the axis of the strain ? Also 
what is the maximum unit shear in 
the bar ? 

30. Stress ellipse. Suppose 
that an elementary triangular 
prism is cut out of a body sub- 
jected to planar strain, so that 
two sides of the prism coincide 
with the principal directions. 
Then, by Article 27, the shears 
in these two sides are zero. Now 
let the axes of coordinates be drawn in the principal directions, and 
resolve the stress acting on the inclined face of the prism into 




FIG. 14 



RELATIONS BETWEEN STRESS AND DEFORMATION 27 

components parallel to the axes instead of into normal and shearing 
stresses as heretofore. Then, from Fig. 14, if dF denotes the area of 
the inclined face, the conditions of equilibrium are 

p' x dF = p x dF sin a, 
p' y dF = pydF cos a ; 



whence 



P 



2 = sin#, 



= cos a. 



Squaring and adding, 



K K 



rf_l 

~ A J 



which is the equation of an ellipse with semi-axes p x and p y) the 
coordinates of any point on the ellipse being p r x and p' y . Conse- 
quently, if the stress acting on the inclined face of the prism is 
calculated for all values of a, and 
these stresses are represented in 
magnitude and direction by lines 
radiating from a common center, 
the locus of the ends of these 
lines will be an ellipse called the 
stress ellipse (Fig. 15). 

31. Simple shear. If a body is 
compressed in one direction and 
equally elongated in a direction at 
right angles to the first, the strain is planar. In this case, if the axes 
are drawn in the principal directions, q = 0, p x = p y , and the stress 
ellipse becomes the circle p' 2 + p'y = p 2 x . 

Moreover, the normal stress in the planes of maximum or mini- 
mum shear is zero; for by substituting in equation (2) the values 
of sin 2 a and cos 2 a obtained from equation (8), the normal stress 

7) ~\~ D 

in the planes of maximum or minimum shear is found to be **- K > 
and this is zero since p x = p y . 

Substituting q = and p x = p y in equation (9), Article 28, the 
maximum or minimum value of the shear in the present case is 




FIG. 15 



[ ruax 

miii 



28 



STRENGTH OF MATERIALS 



that is to say, the intensity of the shear in the planes of zero normal 
stress is equal to the maximum value of the normal stress. 

To give a geometrical represen- 
tation of the conditions of the 
problem, suppose a small cube cut 
out of the body with its faces 
inclined at 45 to the principal 
directions. Then the only stresses 
acting on the inclined faces of 
this cube are shears equal in 
amount to the principal stresses. 
The strain in this case is called 




FIG. 16 



simple shear. 

Conversely, if a small cube is 
subjected to simple shear, as indi- 
cated in Fig. 17, tensile stresses equal in amount to this shear occur 
in the diagonal plane AC of the cube, and compressive stresses of 
like amount in the diagonal plane BD. 



D 



Problem 42. The steel propeller shaft of a 
steamship is subjected to a shearing stress of 
10,000 Ib. /in. 2 Find the maximum tensile stress 
in the shaft. 



32. Coefficient of expansion. Consider 
an infinitesimal prism of dimensions dx, 
dy, dz, and suppose that under strain 
these dimensions become dx + s x dx, 
dy + s y dy, dz + s z dz, where s x) s y , s z are 

the unit deformations in the directions of the edges of the prism. 
Then the volume of the prism becomes 

V+ dV=(dx + s x dx) (dy + s y dy] (dz + s z dz), 
or, neglecting infinitesimals of an order higher than the first, 

V+ dY = (1 + s x + s y + s z )dxdydz. 
Let K = s + s + s 



FIG. 17 



due to the strain is 



Then the change in the volume of the prism 
d V Kdxdydz. 



RELATIONS BETWEEN STRESS AND DEFORMATION 29 

For this reason K is called the coefficient of cubical expansion (or 
contraction) of the body. 

From this definition it is evident that for temperature stresses the 
coefficient of cubical expansion is three times the coefficient of linear 
expansion. 

From Article 9, for linear tensile strain, 



m 



Consequently, in this case, 

Jf <j X X 

-*i- ~ & ~* 



m 2 
m 



E 



Since the prism is certainly not decreased in volume by a tensile 
strain, K cannot be negative and therefore m 2 > 0, or m > 2. If 
m = 2, K = 0, which means 
that the body is incompressi- 
ble. Therefore 2 is the lower 
limit of Poisson's constant. 

33. Modulus of elasticity 
of shear. In an. elementary 
prism subjected to simple 
shear an angular deformation 

occurs, as shown in Fig. 18. FlG lg 

Let the angle of deformation 

</> be expressed in circular measure. Then, for materials which con- 
form to Hooke's law, 




7 



where G is a constant called the modulus of elasticity of shear, or 
modulus of rigidity. Since the angle </>, expressed in circular measure, 
is an abstract number, G must have the dimensions of q, and can 
therefore be expressed in lb./in. 2 , as in the case of Young's modulus. 
Tabulated values of the modulus of elasticity of shear and ultimate 
shearing strength for various substances are given in Table I. 

Problem 43. A f-in. wrought-iron bolt has a diameter of .62 in. at base of 
thread, with a nut f in. thick. What force acting on the nut will strip the thread 
off the bolt ? 



30 



STRENGTH OF MATERIALS 



Problem 44. What force will pull the head off the bolt in Problem 43, if the 
head is of the same thickness as the nut ? 

Problem 45. A f-in. rivet connects two plates which transmit a tension of 
2500 Ib. Assuming that the shear is uniformly distributed over the cross section 
of the rivet, find the unit shear on the rivet. 

Problem 46. An eyebar is designed to carry a load of 15 tons. What^must 
be the size of the pin to be safe against shear ? 

NOTE. Consider the pin in double shear, and assume that this shear is uniformly 
distributed over the cross section of the pin. 

34. Relation between the elastic constants. Suppose a cube is 
subjected to compressive stress on one pair of opposite faces and 



-Px 



dx 



tensile stress on another pair 
of opposite faces. Then, if the 
axes of X and Y are drawn 
in the direction of the strain, 
p x = p y ', and the strain is 
_> one of simple shear, as ex- 
plained in Article 31. 

Let x denote the length 
of an edge of the cube before 
strain. Under the strain the 
cube becomes a parallelepi- 
ped, its increase in length 
in the direction of the X-axis, due to the tensile stress p x , being 

f; and its increase in length in this direction, due to the com- 

Px * 




D 



FIG. 19 



pressive stress p x , being 






Therefore, if dx denotes the total increase in length in the direc- 

tion of the X-axis, 

xp 
dx -- 



xp x 
-*- 



or, since p x = q, 



E mE 

m+l 

dx = xq. 
mE * 



By reason of the strain the angle between the diagonals is increased 
by an amount </>, and therefore the angle between a diagonal and a 

side is increased by From the right triangle ABC (Fig. 19), 

* This assumes that the modulus of elasticity is the same for tension as for com- 
pression. 



RELATIONS BETWEEN STEESS AND DEFORMATION 31 

(IT 6\ x 4- dx 
tan f 4- ~ = 



dx 

\ / 

From trigonometry, 



- tan i 



Since </> is assumed to be very small, tan^ = , approximately, 
and therefore 




2dx 
whence <f> = - = 



x mE 



By definition, G = Therefore 

9 






wliich expresses the relation between the elastic constants G, E, 
and m. 

Problem 47. Fiom the values of G and 22, given in Article 22, determine the 
value of m for cast iron. 

35. Measure of strain. In general, the unit deformation s is taken 
as the measure of a strain. The calculation of s, however, involves 
a knowledge of the modulus of elasticity E, and for many materials 
the latter is difficult to determine. To obviate this difficulty, any 
given strain may be compared with a linear strain which is pro- 
duced by a unit stress equal to the maximum allowable unit stress. 
The stress which would produce this linear strain is called the 
equivalent stress. 

To illustrate the application of this method, consider a planar 
strain in which p l and p 2 denote the principal stresses and s lt s 2 
the corresponding unit deformations. Then, by Hooke's law, the 



32 STRENGTH OF MATERIALS 

stress p l acting alone would produce a unit deformation in the direc- 
tion in which it acts of amount s x = > and also a lateral unit defor- 

o /vj 

mation of th this amount, namely or -^- Similarly, the stress p 
m m mE 

acting alone would produce a unit deformation in its own direction of 

fY\ Q 

amount s. = > and a deformation at right angles of amount or 
E m 

-*-2- Hence the total deformation in the direction in which p l acts, 
say s x , is 



and similarly the total deformation in the direction in which p 2 acts is 

< 13 > = 



Now let p e denote the linear stress which, acting alone, would 
produce the same unit deformation s x or s y ; that is to say, p e is the 
equivalent linear stress which would have the same effect so far as 
deformation is concerned as the combined effect of p l and p z . Then 

s x = (or s = )' and equating these values of s x and s y to those 
E \ E) 

given by equations (12) and (13) above, we have 

(14) P* = Pi P* or p e = p 2 p { . 

lit/ m/ 

The value of the equivalent stress can thus be calculated directly 
from the two principal stresses. In order that the strain be safe, the 
greater of the two values of p e found from equation (14) must not 
exceed the maximum allowable unit stress. 

In the case of simple shear (Article 31) the principal stresses are 
equal in amount to the shear but of opposite sign ; that is, 

Pi= + 2> P 2 =-Q' 

Therefore, inserting these values in equation (14) we have in this case 

1 1 m -f 1 



DELATIONS BETWEEN STRESS AND DEFOKMATION 33 



m 

= 



If, then, the working stress in tension or compression is substituted 
for p e) the allowable shear is given by this relation. 

Problem 48. Find the value of the equivalent stress in Problem 39, and compare 
it with the principal stresses. 

36. Combined bending and torsion. One of the most important 
applications of the preceding paragraph is to the calculation of the 
equivalent stress in a beam subjected simultaneously to bending and 
torsion. 

Let the axis of X be drawn in the direction of the axis of the 
beam. Then on any cross section of the beam there will be a normal 
stress p x due to bending, and a shearing stress q due to torsion, 
while the stress between adjacent longitudinal fibers is zero ; that is, 
p = 0. Therefore, from equation (7), the principal stresses are 



P l = i (P, + 4 f + P% P*=\ (P* ~ 4 <? + Pi)- 
Consequently, from equation (14), the equivalent stress is 
/K\ m 1 , tn + 1 r-L 5 r 

^i^r^-i^ 4 * +^- 

The sign between the terms depends on which of the two values 
for p e in equation (14) is chosen. Evidently that sign should be 
chosen which will give the most unfavorable value of p e . Thus on 
the tension side of a shaft subjected to combined bending and torsion 
the positive sign should be chosen, and on the compression side the 
negative sign. 

If m = 3^, which is the best approximate value to use in general, 
equation (15) becomes 



Many engineers, however, are accustomed to assume .25 for Poisson's 
ratio, making m = 4. The reason for using this value is probably that 
the modulus of rigidity G for most materials is roughly equal to AE\ 



34 STRENGTH OF MATERIALS 

which by equation (11) is equivalent to assuming m = 4. For this 
value of m equation (15) becomes 



Problem 49. A round steel shaft used for transmitting power bears a trans- 
verse load. At the most dangerous section the normal stress due to bending is 
5000 lb./in. 2 , and the shear due to torsion is 8000 lb./in. 2 Calculate the intensity 
of the equivalent stress. 

EXERCISES ON CHAPTER II 

Problem 50. In a boiler plate the tensile stress in the direction of the axis of 
the shell is 2 tons per square inch, and the hoop stress is 4 tons per square inch. 
Calculate the equivalent linear tensile stress. 

Problem 51. At a point in strained material the principal stresses are 0, 
9000 lb./in. 2 tensile, and 5000 lb./in. 2 compressive. Find the intensity and direc- 
tion of the resultant stress on a plane inclined 45 to the axis of the tensile stress 
and perpendicular to the plane which has no stress. 

Problem 52. At a point in the cross section of a girder there is a compressive 
stress of 5 tons/in. 2 normal to the cross section, and a shearing stress of 3 tons/in. 2 
in the plane of the section. Find the directions and amounts of the principal 
stresses. 

Problem 53. At a certain point in a shaft there is a shearing stress of 5000 lb./in. 2 
in the plane of the cross section, and a tensile stress of 3000 lb./in. 2 parallel to the 
axis of the shaft. Find the direction and intensity of the maximum shear. 

Problem 54. Solve Problem 51 graphically by drawing the stress ellipse to scale 
and scaling off the required stress. 

Problem 55. In a shaft used for transmitting power the maximum shearing 
stress, arising from torsional strain, is 5000 lb./in. 2 Find the normal, or bending, 
stress it can also carry if the working stress is limited to 10,000 lb./in. 2 for tension 
or compression, and to 8000 lb./in. 2 for shear. 



CHAPTER III 

ANALYSIS OF STRESS IN BEAMS 

37. System of equivalent forces. The theory of beams deals, in 
general, with the stresses produced in a prismatic body by a set of 
external forces in static equilibrium. Ordinarily these forces all lie 
m one plane ; in this case it is proved in mechanics that they can be 
replaced by a single force acting at any given point in this plane, 
and a moment. To balance this equivalent system of external forces, 
the stresses acting on any cross section of the beam must also consist 
of a single force and a moment, the point of application of this single 
force being conveniently chosen as the 
center of gravity of the cross section. 

The following special cases are of fre- 
quent occurrence. 

If the moment is zero and the single 
force through the center of gravity of a 
cross section acts in the direction of the 
axis of the beam, the strain is simple tension 
or compression ; if it is perpendicular to the axis of the beam, the strain 
is simple shear. 

If the single force is zero and the plane of the moment passes 
through the axis of the beam, pure bending strain occurs ; if the single 
force is zero and the plane of the moment is perpendicular to the 
axis of the beam, a twisting strain called torsion is produced. These 
two cases are illustrated in Fig. 20, A and B. 

If the plane of the moment forms an arbitrary angle with the axis 
of the beam, the moment can be resolved into two components whose 
planes are parallel and perpendicular respectively to the axis of the 
beam. In this case the strain consists of combined bending and torsion. 

If the single force through the center of gravity is inclined to the 
axis of the beam, it can be resolved into two components, one in the 

35 




36 STRENGTH OF MATEEIALS 

direction of the axis, called the axial loading, and the other perpen- 
dicular to the axis, called the shear. 

38. Common theory of flexure. In the majority of practical cases 
of flexure (or bending) of beams, the external forces acting on the 
beam all lie in one plane through its axis and are perpendicular to 
this axis. The single force through the center of gravity of any cross 
section is then perpendicular to the axis of the beam, and the plane of 
the moment passes through this axis. The theory based on the assump- 
tion of this condition of strain is called the common theory of flexure.* 

39. Bernoulli's assumption. In order to obtain a starting point for 
the analysis of stress in beams, the arbitrary assumption is made that 

a cross section of the learn which was plane 
he/ore flexure remains plane after flexure. 
This assumption was first made by Bernoulli, 
and since his time has formed the basis for 
all investigations in the theory of beams. f 

40. Curvature due to bending moment. 
The effect of the external moment on a beam 
originally straight is to cause its axis to be- 
come bent into a curve, called the elastic curve. 
Since, by Bernoulli's assumption, any cross 
section of the beam remains identical with 
itself during deformation, any two consecu- 
tive cross sections of the beam which, .were 

perpendicular to its axis before flexure will remain perpendicular to 
it after flexure, and will therefore intersect in a center of curvature 
of the elastic curve, as shown in Fig. 21. 

The fibers of the beam between these two cross sections were origi- 
nally of the same length. After flexure, however, it will be found that 
the fibers on the convex side have been lengthened by a certain amount 
AB, while those on the concave side have been shortened by an amount 

* The common theory of flexure also includes the following assumptions : (1) the as- 
sumption that Hooke's law is true (Arts. 8 and 11) ; (2) the assumption that plane sections 
remain plane (Art. 39) ; (3) the neglect of vertical shear deformation (Arts.J>8_and 69) ; 
(4) the assumption that dl is equal to dx ; (5) the assumption that the compressive modu- 
lus is equal to the tensile modulus of elasticity ; (6) the neglect of conjugate effect from 
the transverse compression (Art. 9) . 

t St. Venant has shown that Bernoulli's assumption is rigorously true only for certain 
forms of cross section. For materials which conform to Hooke's law, however, it is 
sufficiently exact to assure results approximately correct. 




ANALYSIS OF STRESS IN BEAMS 37 

CD.* Between these two there must lie a strip of fibers which are 
neither lengthened nor shortened. The horizontal line in which this 
strip intersects any cross section is called the neutral axis of the section. 

41. Consequence of Bernoulli's assumption. From Fig. 21 it is 
evident that, as a consequence of Bernoulli's assumption, the length- 
ening or shortening of any longitudinal fiber is proportional to its 
distance from the neutral axis. But, by Hooke's law, the stress is 
proportional to the deformation produced. Therefore the stress on 
any longitudinal fiber is likewise proportional to its distance from 
the neutral axis. Navier was the first to deduce this result from 
Bernoulli's assumption. 

If, then, the stresses are plotted for every point of a vertical strip 
MN (Fig. 22), their ends will all lie in a straight line, and conse- 
quently this distribution of stress is called the 
straight-line law. 

42. Result of straight-line law. In rectan- 
gular coordinates let the axis of Z coincide 
with the neutral axis, and the axis of Y be 
perpendicular to it and in the plane of the 

cross section. Then if the normal stress at FlG - 22 

the distance y from the neutral axis is denoted by p, and that at a 

distance y Q is denoted by p Q) from the straight-line law, 

' (16) *=. 

PO y 

Since in order to equilibrate the external bending moment the normal 
stresses must also form a moment, the sum of the compressive stresses 
must equal the sum of the tensile stresses. Therefore, since the tensile 
and compressive stresses are of opposite sign, the algebraic sum of 
all the normal stresses acting on the section must be zero, that is to 

say, I pdF = 0, where dF is the infinitesimal area on which p acts. 
Inserting the value of p from (16), 

* This can be shown experimentally by placing two thin steel strips in longitudinal 
grooves in a wooden beam, one on the upper side and the other on the lower side, so that 
the strips are free to slide longitudinally but are otherwise fixed. If the strips are of 
the same length as the beam before bending, it will be found that after bending the upper 
strip projects beyond the ends of the beam, while the lower strip does not reach the ends. 
Experiments of this kind have been made by Morin and Tresca. See Unwin, The Testing 
of Materials of Construction, p. 36. 




38 STRENGTH OF MATERIALS 

/Po 
qj 

and therefore 

/ 

= 0. 



But the distance of the center of gravity of the section from the axis 
of Z (or neutral axis) is given by 



fydJF 

- - 
f 



y = 

a 



Therefore, since / yd F = 0, y must be zero, and consequently the 

neutral axis passes through the center of gravity of the section. 

43. Moment of inertia. For equilibrium, the moment of the nor- 
mal stresses acting on any cross section must equal the moment of 
the external forces at this section. Therefore, if M denotes the 
moment of the external forces, or external bending moment, as it is 
called, 

ipydF = M, 
or, from (16), 



The integral J y*dF depends only on the form of the cross section, 

and is called the moment of inertia of the cross section with respect to 
the neutral axis. 

Let the moment of inertia be denoted by Z Then 

I- 

and, consequently, 

%o 

(17) Po = -J-' 

This formula gives the intensity of the normal stress p at the distance 
2/ from the neutral axis, due to an external bending moment M. If 



ANALYSIS OF STRESS IN BEAMS 



39 



p denotes the stress on the extreme fiber and e denotes the distance 
of this fiber from the neutral axis, then, from (17), 



(18) 



p = 



Me 



Equation (18) gives the maximum normal stress on any cross section 
of a beam, and is the fundamental formula in the common theory of 
flexure. 

Problem 56. Find the moment of inertia of a rectangle of height h and breadth 
b about a gravity axis * parallel to its base. 



Solution. 



T\ 


\ 


\ 


\ 




i 


1 




J |_ 


i 


j._ 


~T 


1 


1 


I 




I 
1 


1 



Problem 57. Find the moment of inertia of a triangle of base 6 and altitude h 
about a gravity axis parallel to its base. 

Problem 58. Find the moment of inertia of 
a circle of diameter d about a gravity axis. 

Problem 59. The external moment acting 
on a rectangular section 12 in. deep and 4 in. 
wide is 30,000 ft. Ib. Find the stress on the 
extreme fiber. 

Solution. M= 30,000ft. Ib. = 360,000 in. Ib., 

!, = = 576 in.*. 
12 

.-. p = = 3750 Ib./inA i 

1 FIG. 23 

44. Moment of resistance. The moment of resistance is defined as 
the moment of the internal stresses which balances the external moment 
M . According to this definition the moment of resistance is simply 



vl 
since -M. Therefore, if p is the maximum allowable unit stress 

e vl 

for any material, the moment of resistance determines the 

e 

maximum external bending moment which can be safely carried by 
a beam of this material. 



*In what follows, "gravity axis" will be used as an abbreviation for " axis through 
the center of gravity." 



40 STRENGTH OF MATEKIALS 

For instance, consider an oak beam 8 in. deep and 4 in. wide. From Table I, 
the ultimate compressive strength for timber may be tal^en as 7000 lb./in. 2 , and 
the ultimate tensile strength as 10,000 lb./in. 2 Therefore, using a factor of safety 
of 8, the safe unit stress is p = 875 lb./in. 2 For the beam under consideration 
I = 170.7 in. 4 and e = 4 in. Consequently, the maximum bending moment which 
the beam can be expected to carry safely is 37,340 in. lb., or 3112 ft. Ib. 

Problem 60. Find the moment of resistance of a circular cast-iron beam 6 in. 
in diameter. 

Problem 61. Find the moment of resistance of a Carnegie steel I-beam, No. B 1, 
weighing 80 Ib./f t. 

Problem 62. Compare the moments of resistance of a rectangular beam 
8 in. x 14 in. in cross section, when placed on edge and when placed on its side. 

45. Section modulus. Iii Article 43 the moment of inertia was 
defined as the integral 

/= CfdF. 

From this definition it is apparent that the moment of inertia de- 
pends for its value solely on the form of the cross section. Since it 
is independent of all other considerations, it may therefore be called 
the shape factor in the strength of materials. 

Since e denotes the distance of the extreme fiber of a beam from 

the neutral axis, the ratio - is also a function of the shape of the 

e 

cross section, and for this reason is called the section modulus. Let 

the section modulus be denoted by S. Then S=- t and the expres- 
sion for the moment of resistance becomes 

M = pS. 

This expresses the fact that the strength of a beam depends jointly on 
the form of cross section and the ultimate strength of the material. 

Problem 63. Find the section moduli for the sections given in Problems 56, 57, 
and 58 respectively. 

Problem 64. Compare the section moduli for a rectangle 10 in. high and 4 in. 
wide, and for one 4 in. high and 10 in. wide. 

46. Theorems on the moment of inertia. The following is a sum- 
mary of the most useful theorems concerning the moment of inertia. 
The proofs can be found in any standard text-book on mechanics. 

(A) Let I g denote the moment of inertia of any cross section with 
respect to a gravity axis (see footnote, p. 39), I n the moment of inertia 



ANALYSIS OF STRESS IN BEAMS 



41 




FIG. 24 



of the same section with respect to any parallel axis, c the distance 
between the two axes, and F the area of the cross section. Then 

(19) I n = I 9 + F <?- 

(B) Every section has two axes through its center of gravity, called 
principal axes, such that for one of these the moment of inertia is 

a maximum, and for the other is a ^, ->^ / n 

minimum. Let the principal axes be 

taken for the axes of Y and Z re- 
spectively. Then if I y and I z denote 
the moments of inertia of the section 
with respect to these axes, and I a 
denotes the moment of inertia with 
respect to an axis inclined at an angle a to the axis of Z, 

(20) I a = I x cos 2 a; -f / sin 2 a. * 

(C) The moment of inertia of a compound section about any axis is 
equal to the sum of the moments of inertia about this axis of the 
various parts of which the compound section is composed. 

(D) The moment of inertia of any section with respect to an axis 
through its center of gravity and perpendicular to its plane is called 

the polar moment of inertia. The polar 
moment of inertia is defined by the 
equation 

4- 

where r is the distance of the infini- 
tesimal area dF from the center of 
gravity of the section. 

Since r 2 = y 2 " + z*, 




FIG. 25 



C r 2 dF = Cy*dF+ Cz?dF, whence 



(21) 



(E) Let Jj and / 2 denote the moments of inertia of any section with 
respect to its principal axes. Then I p = I I + / 2 , and, consequently, 

* If the axes of Y and Z are not principal axes, then 

la = Iz cos 2 a + Iy sin 2 a ffyz dy dz. 



42 



STRENGTH OF MATERIALS 



(22) 

that is to say, the sum of the moments of inertia with respect to any 
two rectangular axes in the plane of the section is constant. 

(F) The numerical value of the moment of inertia is expressed as 

the fourth power of a unit of length. Therefore the quantity is 

F 

~^ the square of a length called the radius of 

gyration, and will be denoted by t. The 
radius of gyration is thus defined by the 
_ equation 



(33) 



F IG> 26 that is to say, the square of the radius 

of gyration is the mean of the squares 
of the distances of all the elements of the figure from the axis. 

The meaning to be attached to the radius of gyration is that if 
the total area of the figure was concentrated in a single point at a 
distance t from the axis, the moment of inertia of this single particle 
about this axis would be equal to the given moment of inertia. 

Problem 65. Find the moment of inertia of the rectangle in Problem 56 about 
its base, and also the corresponding radius of gyration. 




T 

Solutwn - ^ 



bh* 



Problem 66. Find the moment of inertia of 
the above rectangle about a gravity axis inclined 
at an angle of 30 to its base. 

Problem 67. Find the moment of inertia of a 
rectangular strip, such as that shown in Fig. 26, 
about a gravity axis parallel to its base. 

Problem 68. Prove that the moment of inertia 
of a T-shape, such as that shown in Fig. 27, about 
a gravity axis parallel to the base Is given by the 
expression 



6' i 



t 



6 

FIG. 27 



Problem 69. Find the polar moment of inertia and radius of gyration of a circle 
of diameter d about an axis through its center. 



ANALYSIS OF STRESS IN BEAMS 



43 



47. Graphical method of finding the moment of inertia. If the 

boundary of a given cross section is not composed of simple curves 
such as straight lines and circles, it is often difficult to find the 
moment of inertia by means of the calculus. When such difficulties 
.arise the following graphical method may be used to advantage. 

To explain the method consider a particular case, such as the rail 
shape shown in Fig. 28, and suppose that it is required to find the 
center of gravity of the section, and also its moment of inertia about 
a gravity axis perpendicular to the web. The first step is to draw two 
lines, AB and CD, par- , 

allel to the required ^ ^ H 

gravity axis, at any k | g"-l *\ 

convenient distance f *^T T~". 

apart, say /. 

If the section is sym- 
metrical about any axis, 
such as Y in the fig- 
ure, it is sufficient to 
consider the portion 
on either side of this 
axis, say the part on 
the right of Y in the ^ 
present case. 

Now suppose that 

the cross section is di- FIG 

vided into narrow strips 
parallel to AB and CD ; let z denote the length of one of these strips, 




and dy its width, 
such that 



Then, if for each value of z a length z r is found, 



any point P on the boundary of the original section, with coordinates 
z and y, will be transformed into a point P' with coordinates z 1 and y. 
Suppose this process is carried out for a sufficient number of points, 
and that the points P' so obtained are joined by a curve, as shown 
by the dotted line in Fig. 28. Let F denote the area of the original 
curve and F 1 the area of the transformed curve, both of which can 



44 STRENGTH OF MATERIALS 

easily be measured by means of a planimeter. Also let N denote the 
static moment of the original section with respect to the line AB, 
where the static moment an area with respect to any axis is 
defined by the integral 



in which y is the distance of an infinitesimal area dF from the given 
axis. The static moment is thus equal to the area of the section 
multiplied by the distance of its center of gravity from the given 
axis. Then 



N = CydF = Cyzdy = I Cz'dy = IF'. 



But, from the above definition, 

N=cF, 

where c is the distance of the center of gravity of the original sec- 
tion from the line AB. Therefore cF = IF' whence 



which determines the position of the center of gravity. 

To find the moment of inertia, make a second transformation by 
constructing for each z' a value z fr , such that 

*=*&. 

Then the points P' on the first transformed curve are transformed 
into a series of points P rf on another curve, shown by the broken 
line in Fig. 28. Let the area of this second curve be denoted by F". 

Then, since z" =dj> and z' = z j > we have z" = z ^- - Consequently, 
I I I 



= Cy*dF = Cfzdy = l z Cz"dy = l*F", 



which gives the moment of inertia of the original section with respect 
to the line AB. 



ANALYSIS OF STRESS IN BEAMS 



45 



If the moment of inertia I g with respect to a gravity axis is required, 
then, since by Article 46 (A), /= I g + c*F, we have I g = l<?F\ and 
hence, by substituting the values of / and c from the above, 



! = 



The above method is due to Nehrs, and furnishes an easy method 
of calculating the moment of inertia of any cross section by simply 
measuring the area F of the original section and the area F', F" of 
the transformed sections by means of a planimeter, and then substi- 
tuting these values in the above formulas. 

48. Moment of inertia of non-homogeneous sections. The stand- 

Me 

ard formula for calculating the stress in beams, p = , assumes 

that the material of which the .beam is composed is homogeneous 
throughout. If, then, a beam is com- i r 

posed of two different materials, such, 
for instance, as concrete and steel, it is 
necessary to modify this formula some- 
what before applying it. 

To exemplify this, consider a rectan- 
gular concrete beam, reenforced by steel 
rods near the bottom, as shown in cross 
section in Fig. 29. Let p c and p s denote 
the stresses on a fiber of concrete and 
of steel respectively, at the same distance y from the neutral axis, and 
let E t and E g denote the moduli of elasticity for concrete and steel. 
Then, by Hooke's law, 







FIG. 29 



whence 



jn > 

A, 



, 
P.-JP, 



Therefore, if dF is an infinitesimal area of steel at the distance y 
from the neutral axis, the moment of the stress acting on this area is 



K f 



46 



STRENGTH OF MATERIALS 



Consequently, the intensity of the fiber stress can be considered to 
vary directly as its distance from the neutral axis over the entire 
cross section of the beam, provided the area of the steel is increased 

7f 

in the ratio '- If, then, the depth is kept constant, the breadth 

E c 

must be increased in this ratio, and the cross section thus obtained 









1 








e 






o 


L * 




" 






1 






\ 




















FIG 


.30 





will appear as shown in Fig. 30. Therefore, if I c denotes the moment 
of inertia of this modified section, the stress in the extreme fiber is 
given by the formula 



p = 



Me 



i* 8 





f 1 
1 
If, 

7.69 











12 25" 1 

*_ 






" 


z_ 


^ 1 




















1 t 










k 


29.1 





\ 





FIG. 31 



FIG. 32 



Problem 70. A rectangular concrete beam 14 in. deep and 8 in. wide is reenforced 
by two f-in. square steel rods placed 1 in. from the bottom, as shown in Fig. 31. 
Assuming that the ratio of the moduli of elasticity of steel and concrete is 
E s : E c = 15 : 1, find the moment of inertia of a cross section of the beam about a 
gravity axis parallel to the base. 

Solution. Increasing the area of the steel in the rate 15 : 1, it becomes 16.9 in. 2 . 
The area of the concrete included in the same horizontal strip with the steel is 



ANALYSIS OF STRESS IN BEAMS 



47 



4.9 in. 2 . Consequently, the breadth of the lower flange of the equivalent homo- 
geneous section is 

16.9 + 4.9 



.75 



= 29. 1 in. 



The distance of the center of gravity of this equivalent section below the top 
is found to be 7.69 in., and its moment of inertia about the gravity axis OZ is 
2269 in.* (Fig. 32). 

49. Inertia ellipse. Dividing equation (20) by F and expressing 
the result in terms of the radii of gyration by means of equation (23), 



(24) 



tl cos 2 a -f sin 2 a, 



where t y and t e are the radii of gyration with respect to the axes of 
Y and Z respectively, and t a is the radius of gyration with respect to 
a gravity axis inclined at an angle a to the axis of Z. 

Now let I be a length defined by the relation -&-* = I. Then 

It It ^ 

t y = - , t z = - ; and substituting these values of t y and t t in equa- 



tion (24), it becomes 



or, dividing by 



p% 

<J 



7 2 / 

u U/ 



1 = 



This is the equation of an ellipse 
with semi-axes t y and t,, called the 
inertia ellipse, the coordinates of 
any point of the curve being I cos a 
and / sin#. 

By means of the inertia ellipse 
the moment of inertia with re- 
spect to any gravity axis AB (Fig. 
33) can be obtained as follows. ^ 2 

The equation of a tangent to the ellipse + ^ = 1 at the point 




FIG. 33 



(25) 



+ yy'c? - aV = 0. 



48 



STRENGTH OF MATERIALS 



It is proved in analytical geometry that in order to reduce the linear 
equation Az-\-By + C= to the normal form z cos /3-f- y sin ft c = 0, 
it is necessary to divide throughout by V^ 2 + B*. Applying this 
theorem to equation (25), it becomes 



= 0, 




Substituting these values in the expression a 2 cos 2 /3 + b 2 sin 2 /3, it 
becomes 



whence, since ft = a > 

c 2 = a 2 cos 2 /3 -f b 2 sin 2 /3 = a 2 sin 2 a + b 2 cos 2 a. 

Since the semi-axes of the inertia ellipse are a = t y and & = t tt this 
expression becomes 2 , 2 . 2 2 

/ / cir> /y I /^ pnta^/y 

t Olll 14> ~|~ t/ a l^UiS I*, 

or, comparing this expression with equation (24), 



The radius of gyration corresponding to any gravity axis AB can 
therefore be found by drawing a tangent to the 
inertia ellipse parallel to AB, and measuring the 
distance of this tangent from the center. 

Since the inertia ellipse is constructed on the 
principal radii of gyration as semi-axes, it can be 
drawn on all the ordinary forms of cross section, 
and when this is done the method given above 
greatly simplifies the calculation of the moment 
of inertia with respect to any gravity axis which 
FIG. 34 is not a principal axis. 

Problem 71. From the Carnegie handbook of structural steel the principal 
radii of gyration of T-shape, No. 72, size 3 in. by 4 in., are 1.23 in. and .59 in. 
Construct the inertia ellipse (Fig. 34). 




_ I ______ i 



ANALYSIS OF STRESS IN BEAMS 49 

Problem 72. For a Carnegie I-beam, No. B 7, 15 in. deep and weighing 
42 lb./ft., the principal radii of gyration are 5.95 in. for an axis perpendicular to 
web at center, and 1.08 in. for an axis coincident with web at center. Construct 
the inertia ellipse. 

Problem 73. For a Cambria channel, No. C 21, depth of web 7 in., width of 
flanges 2.51 in., thickness of web .63 in., the radius of gyration about an axis per- 
pendicular to the web at center is 2.39 in.; the distance of the center of gravity 
from outside of web is .58 in., and the radius of gyration about an axis through 
the center of gravity parallel with center line of web is .56 in. Construct the 
inertia ellipse. 

Problem 74. In Problems 68, 69, and 70 determine graphically the radii of 
gyration about an axis through the center of gravity and inclined at 30 to the 
major axis of the inertia ellipse. 

50. Vertical reactions and shear. Under the assumptions of the 
common theory of flexure, the external forces acting on a beam all 
lie in the same vertical plane. Therefore, since the beam is assumed 
to be in equilibrium, the sum 
of the reactions of the sup- L 

ports must equal the total [* ^ 1 _ 
load on the beam. 

For instance, consider a 

r> !< - ..... - 7 

simple beam AB of length /, 

which is supported at the F IG . 35 

ends and bears a single con- 

centrated load P at a distance d from A (Fig. 35). Let R^ and R 2 

denote the reactions at A and B respectively. Then, from the above, 

R l + R z = P. 

To find the values of R^ and R z , take moments about either end, say A. 

Then 

RJ, = Pd\ 

whence 



Also, since 



If any cross section of a beam is taken, the stresses acting on this 
section must reduce to a single force and a moment, as explained in 




50 STRENGTH OF MATERIALS 

Article 37. For a simple beam placed horizontally and supporting a 
system of vertical loads, the plane of the moment is perpendicular to 
the plane of the section, and the single force is a vertical shear lying 
in the plane of the section. Therefore, since the portion of the beam 
on either side of the section must be in equilibrium, the vertical 
shear is equal to the algebraic sum of the external forces on either 
side of the section. Thus, if the portion of the beam on the left of 
the section is considered, the vertical shear on the section is equal 
to the reaction of the left support minus the sum of the loads on the 
left of the section. 

Problem 75. A beam 10 ft. long bears a uniform load of 300 Ib./ft. Find the 
vertical shear on a section 4 ft. from the left support. 

Solution. The total load on the beam is 3000 Ib. Therefore, since the load is 
uniform, each reaction is equal to 1500 Ib. The load on the left of the section is 
300 x 4 1200 Ib. Therefore the vertical shear on the section is 1500 1200 = 300 Ib. 

Problem 76. Find the vertical shear at the center and ends of the beam in the 
preceding problem. 

Problem 77. A beam 12 ft. long bears loads of 1, |, and 3 tons at distances of 
2, 6, and 7 ft. respectively from the left support. Find the vertical shear at either 

end of the beam, and also at a 
point between each pair of loads. 



UJ. 



( X 



51. Maximum bending 
moment . The external bend- 
ing moment at any point of 

T> 

2 a beam is denned as the sum 
of the moments, about the 
neutral axis of a cross sec- 
tion through the point, of all 
the external forces on either 



FIG. 36 side of the section. Thus, if 

the portion of the beam on 

the left of the section is considered, the external moment at this point 
is the moment of the reaction of the left support about the neutral 
axis of the. section, minus the sum of the moments of the loads 
between the left support and the section, about the same neutral axis. 

For example, in Fig. 36 the moment of RI about the neutral axis of the section 
mn is BIZ, and the moment of PI about the same axis is PI (x di). Therefore 
the total external moment acting on the section mn is 

M=E l x-P l (x-di). 



ANALYSIS OF STKESS IN BEAMS 51 

As another example, consider a beam of length I bearing a uniform load of 
amount w per unit of length. Then the total load on the beam is wl, and each 
reaction is Therefore the moment at a point distant x from the left support is 

ivl x wx ., 



From this relation it is evident that M is zero f or x = or Z, and attains its maxi- 
mum value for x = - ; that is to say, the bending moment is zero at either end of 
the beam and a maximum at the center. 

From the formula M =pS y given in Article 45, it is evident that 
the maximum value of the stress p occurs where the bending moment 
M is a maximum. Ordinarily the maximum bending moment pro- 
duces a greater strain than the maximum shear ; therefore the section 
at which the maximum moment occurs is called the dangerous section, 
since it is the section at which the material is most severely strained, 
and consequently the one at which rupture may be expected to occur. 

In order to find the maximum bending stress in a beam, the formula 

M = pS is written 

M 



The maximum bending stress is then obtained at once by simply 
dividing the maximum bending moment by the section modulus. 

Problem 78. A rectangular wooden beam 14 ft. long, 4 in. wide, and 9 in. deep 
bears a uniform load of 75 Ib./ft. Find the position and amount of the maximum 
bending moment. 

Problem 79. Find the maximum bending stress in the beam in the preceding 
problem. 

Problem 80. A Cambria I-beam, No. B 33, which weighs 40 Ib./ft., is 15 ft. 
long and bears a single concentrated load of 5 tons at its center. Find the maxi- 
mum bending stress in the beam, taking into account the weight of the beam. 

52. Bending moment and shear diagrams. In general, the bending 
moment and shear vary from point to point along a beam. This 
variation is shown graphically in the following diagrams for several 
different systems of loading. 

(A) Simple beam bearing a single concentrated load P at its center 

(Fig. 37). From symmetry the reactions 7^ and fi 2 are each equal 

-p 
to Let mn be any section of the beam at a distance x from the 

& 

left support, and consider the portion of the beam on the left of this 



52 



STRENGTH OF MATERIALS 




section. Then the moment at mn 
s ^x = x ) and the shear is 



= ). For a section on the 



FIG. 37 



right of the center the bending 
moment is R 2 (l x) and the shear 
is Rg Consequently, the bending 
moment varies as the ordinates of 
a triangle, being zero at either sup- 
port, and attaining a maximum 

PI 
value of at the center, while 

the shear is constant from A to B, 
and also constant, but of opposite 
sign, from B to C. 
The diagrams in Fig. 37 represent these variations in bending 
moment and shear along the beam under the assumed loading. Con- 
sequently, if the ordinates vertically beneath B are laid off to scale 
to represent the bending moment and shear at this point, the bending 

moment and shear at any ^ 7 ^ 

other point D of the beam 
are found at once from the 
diagram by drawing the 
ordinates EF and HK verti- 
cally beneath D. 

(B) Beam bearing a single 
concentrated load P at a dis- 
tance c from one support. 

The reactions in this casj 
are 

_P(l-c) p (W 

M ; 2 



and 



Pc 



Hence the bending moment 




ANALYSIS OF STRESS IN BEAMS 



53 



at a distance x from the 
left support is 



provided x < c, and 
_Pc(l- 



R. 



if x > c. If x = c, each of 
these moments becomes 

Pc(l-c) 



and consequently the bend- 
ing moment and shear dia- 
grams are as shown in 
Fig. 38. 

(C) Seam bearing sev- 
eral separate loads. 




SHEAR 



FIG. 30 



In this case the bending moment diagram is obtained by con- 
structing the diagrams for 
each load separately and 
then adding their ordinates, 
as indicated in Fig. 39. 

(D) Beam bearing a con- 
tinuous uniform load. 

Let the load per unit of 
length be denoted by w. 
Then the total load on the 
beam is wl, and the reac- 
tions are 

wl 



Hence at a distance x from 
the left support the bend- 
ing moment M x is 




54 



STRENGTH OF MATERIALS 



wl 
~2 



The bending moment diagram is therefore a parabola. For x = > 

wl 2 

M x = which is its maximum value. The bending moment and 
8 

shear diagrams are therefore as represented in Fig. 40. 

(E) Beam bear- 
ing uniform load 
over part of the 
span. 

Let the load ex- 
tend over a distance 
c and be of amount 
w per unit of length. 
Then the total load 
is we. The reactions 
of the supports are 
the same as though 
the load was concen- 
trated at its center 
of gravity G. There- 
fore, if d denotes the 
distance of G from 
the left support, 

wc(l-d) 



Also, the bending moment diagrams for the portions AB and CD are 
the same as though the load was concentrated at G, and are there- 
fore the straight lines A'H and D'K, intersecting in the point T 
vertically beneath G (Fig. 41). 

From B to C there is an additional bending moment due to the 
uniform load on this portion of the beam. Thus, if LMN is the para- 
bolic moment diagram for a beam of length LN or c, the ordinates 
to the line HK must be increased by those to the parabola LMN, 
giving as a complete moment diagram the line A'HJKD'. 




ANALYSIS OF STRESS IN BEAMS 



55 



Analytically, if x denotes the distance of any section from the left 
support, the equations of the three portions A'H, HJK, and KD' of 
the moment diagram are 



-"^AB **jtf> 


i 


JLUl 


V/ <, i/O <, U* - 


2' 






1 c\ 


12 




/ 


^ C V 




L 2 / 


we (I 


- d) x 


^r~ 


+ 2J 



for 



^cc? (I x) , 
= R z x = ^- ^ for 



c = = 7 
d -j < x < /. 



Problem 8 1 . Construct the bending moment and shear diagrams for a cantilever * 
bearing a single concentrated load P at the end. 

Problem 82. Construct the bending moment and shear diagrams for a simple 
beam bearing two equal concen- 
trated loads at equal distances 
from the center. 

53. Relation between 
shear and bending moment. 

Consider a beam bearing sev- 

eral concentrated loads P v 

P 2 , etc., at distances d v d z , 

etc., from the left support. Take any section mn at a distance x 

from the left support, and consider the portion of the beam on the 

left of this section. Then if Q denotes the total shear on this section, 





IP, P 2 m 


r , 


A 




j 






/fl 


r7 


_ - - -^j 






tt4 

FIG. 


42 





Also, the bending moment at mn is 



where the summations include all the loads between A and the 
section mn. 



* A cantilever is a beam which is framed into a wall or other support at one end and 
projects outward from this support. 



56 STRENGTH OF MATERIALS 

Differentiating M with respect to x, 



Therefore 



that is to say, the shear at any point of a beam is the first differential 
coefficient of the bending moment at that point. 

If the beam is uniformly loaded, as in (Z>) of the preceding 

IJJ'li 

article, Q = R^ wx and M = R^x --- > from which equation (26) 
results as before. 

From equation (26) it follows that if the bending moment is con- 
stant the shear is zero ; and conversely, if the shear is zero the bend- 

ing moment is constant. But = is the condition that the 

CLOu 

bending moment shall be either a maximum or a minimum. Conse- 
quently, at a point where the bending moment passes through a maxi- 
mum or minimum value the shear is zero ; and conversely. This 
theorem is illustrated by the bending moment and shear diagrams in 
the preceding paragraph. 

54. Designing of beams. In designing beams the problem is to 
find the transverse dimensions of a beam of given length and given 
material, so that it shall bear a given load with safety. 

In order to solve this problem, the formula M = pS is written 

- = S. 
p 

Then, from the given loading, the maximum value of M is determined, 
and by dividing the ultimate strength of the material by the proper 
factor of safety the safe unit stress p becomes known. The quotient 
of these two gives the section modulus of the required section. 

In the handbooks issued by the various structural iron and steel 
companies, the section moduli of all the standard sections are tabu- 
lated. If, then, the beam is to be of a standard shape, its size is 
found by simply looking in the tables for the value of S which corre- 

sponds most closely to the calculated value > the value chosen 

P 



ANALYSIS OF STKESS IN BEAMS 



57 



being equal to or greater than the calculated value in order to insure 
safety. 

If the section of the beam is to be of a shape not listed in the 
handbooks, the dimensions of the section must be found by trial. 
Thus a section of the required shape is assumed, and its section 
modulus calculated from the relation 



If the value of S thus found is too great or too small, the dimensions 
of the section are decreased or increased, and S again calculated. 
Proceeding in this 
way, the dimensions 
of the section are 
changed until a 
value of S is found 
which is approxi- 
mately equal to the 

calculated value 
P 

Problem 83. Design 
a steel I-beam, 10 ft. 
long, to bear a uniform 
load of 1600 lb./ft., neg- FIG. 43 

lecting its own weight. 

Problem 84. A built beam is to be composed, of two steel channels placed on 
edge and connected by latticing. What must be the size of the channels if the 
beam is to be 18 ft. long and bear a load of 10 tons at its center, the factor of 
safety being given as 4 ? 

Problem 85 . Compare the strength of a pile of 10 boards, each 14 ft. long, 1 ft. 
wide, and 1 in. thick, when the boards are piled horizontally, and when they are 
placed close together on edge. 

Problem 86. Design a rectangular wooden cantilever to project 4 ft. from a 
wall and bear a load of 500 Ib. at its end, the factor of safety being 8. 

Problem 87. A rectangular cantilever projects a distance I from a brick wall 
and bears a single concentrated load P at its end. How far must the inner end of 
the cantilever be imbedded in the wall in order that the pressure between this end 
and the wall shall not exceed the crushing strength of the brick ? 

Solution. Let 6 denote the width of the beam and x the distance it extends into 
the wall. For equilibrium the reaction between the beam and the wall must con- 
sist of a vertical force and a moment. If p a denotes the intensity of the vertical 




58 



STRENGTH OF MATERIALS 



stress, and it is assumed to be uniformly distributed over the area &c, p a bx P; 

p 
whence p a = (see Fig. 43, a). 

Similarly, let p b denote the maximum intensity of the stress forming the stress 
couple. Then, taking moments about the center C of the portion AB, since the 
stress forming the couple is also distributed over the area 6x, we have 



bx* 



and 



Me 



Therefore, substituting in the formula p = , we have 



(-1)1 



12 



6P 



Pmax = Pb Pa = 
min 



H) 



= 2P / 3A 

" bx \ + x) 



Consequently, 
whence 

and 



As a numerical example of the above, let I = 5 ft. = 60 in., P = 200 lb., 6 = 4 in., 
and p = 600 lb./in. 2 (for ordinary brick work). Solving the above equation by the 

formula for quadratics, 



6<>^ 


^ 


^F/ 


n 


2 P V4 P 2 + 6 &pPi 


<2L. i 


} ^p 


S^$p r /Z 


1 


&p 




* ~% 

^ ^ 


Ur^ 


\ 


whence, by substituting the above values, 
x- 5.6 in. 


j_ 


A B 


c y*^ 

R 


I 


|55. Distribution of shear over 
rectangular cross section Con- 


^''' 

S"* 


S,^"' 


.*'' ^S 


GJ 


sider a cross section of a rectan- 

EmlflT bpflTTI flf" f> rH<sf pn<"> T frrvm 




FIG. 44 


H 




the left support, as MNRS in Fig. 44, 
and let P be a point in this cross 



Article 43, the unit normal stress at P is p = K 



If the cross 



* Bach, Elasticitat u. FestigkeitsleJire, p. 430. 

t For a brief course in the Strength of Materials the remainder of this chapter may 
be omitted. 



ANALYSIS OF STRESS IN BEAMS 59 

section is moved from this position parallel to itself a distance dx, 
say to the position EFGH in the figure, the rate of change of p with 
respect to x is 

/ 27 ^ dp _dM y _ y 

to dZ I I Q ' 

The difference between the normal stresses acting on these two 
adjacent cross sections tends to shove the point P in a direction 
parallel to the axis of the beam, and this tendency is resisted by 
a shearing stress of intensity q at P, also parallel to the axis of 

the beam. Therefore, since the resultant normal stress on the area 
* 

r 2 

BCEF is I dp-dF, and the resultant shearing stress on the area 

*) c 

ABCD is qbdx, h _ 

/2 
dp - dF = qbdx. 

Substituting the value of dp from equation (27), 



whence h 

(28) q = Q CydF. 

Formula (28) applies to any cross section bounded by parallel sides. 

In Article 23 it was proved that whenever a shearing stress acts 
along any plane in an elastic solid, there is always another shearing 
stress of equal intensity acting at the same point in a plane at right 
angles to the first. Consequently, formula (28) also gives the intensity 
of the stress at any point P in a direction perpendicular to the neutral 
axis of the section. 

For a rectangular cross section 



and hence 

(29) 



i8 a 



60 



STKENGTH OF MATEKIALS 



From equation (29), it is evident that for rectangular sections the 

shear is zero at the top and bottom of the beam ( where c = - ) and 

\ 2/ 

increases toward the center as the ordinates to a parabola. For c = 0, 

q attains its maximum value, namely, q = ~- (Fig. 45). At the top 

and bottom where the normal bending stress is greatest the shear is 
zero, and at the center where the normal stress is zero the shear is a 
maximum. 




1 




FIG. 45 

Since the area of the parabola 
ABC is %hq, the average stress is 
I hq/h = | q, and consequently the 
maximum unit stress q is | average 
unit stress. 

56. Distribution of shear over 
circular cross section. For a rec- 
tangular cross section the shear parallel to the neutral axis is zero, 
but for a circular cross section this is not the case. Let Fig. 46 rep- 
resent a circular cross section, say the cross section of a rivet sub- 
jected to a vertical shear, and let it be required to find the direction 
and intensity of the shear at the extremity N of a horizontal line 
MN. If the stress at N has a normal component, that is, a compo- 
nent in the direction ON, it must have a component of equal amount 
through N perpendicular to the plane of the cross section, that is, 
in the direction of the axis of the rivet (Article 23). Consequently, 
since the rivet receives no stress in the direction of its axis, the stress 
at N can have no normal component and is therefore tangential. 

Similarly, the stress at M is tangential, and since the line MN is 
horizontal, the tangents at M and N must meet at some point B on 
the vertical diameter, which is taken for the F-axis. The stress at 
any point K on the F-axis must act in the direction of this axis, and 



ANALYSIS OF STRESS IN BEAMS 61 

therefore also pass through B. For. any other point of MN it is 
approximately correct to assume that the direction of the stress also 
passes through B. 

Therefore, in order to determine the direction and intensity of the 
shear at any point of a circular cross section, a chord is drawn through 
the point perpendicular to the direction of the shear and tangents 
drawn at its extremities, thus determining a point such as B in 
Fig. 46. Assuming the axes as in Fig. 46, the vertical shear acting 
at the point is then calculated by formula (28), where, in the present 
case, b is the length of the chord and the integral is extended over 
the segment above the chord. The horizontal component of the shear 
is then determined by the condition that the resultant of these two 
components must pass through B. 

The amount of the component and resultant shears acting at any 
point can be calculated as follows. 

For a strip parallel to the Z-axis, dF = zdy, and z = Vr 2 y 2 . 
Therefore 



The vertical component of the shear is, therefore, 

^ = &I\12 

Let KB and KN, Fig. 46, represent in magnitude and direction 
the vertical and horizontal components of the shear acting at N. Then, 
from the similar triangles KNB and KNO, 



KB KN 
whence 



_ _ 

~ ~' 



Since BN* = BK* + KN*, the resultant shear at N is 



62 



STRENGTH OF MATERIALS 



or, since - + A 2 



In this equation q is proportional to 5, and hence the maximum 
value of q is at the center where b = 2 r. Hence 

(/max = 




E 
FIG. 47 



The maximum unit shear on a circular cross section is therefore 

equal to of its average value. 

57. Cases in which shear is of especial importance. In Article 53 

it was shown that at points where the normal bending stress is a 
j_ maximum the shear is zero. 

For this reason it is usu- 
ally sufficient to dimension 
a beam so as to carry the 
maximum bending stress 
safely without regard to 
the shear. However, in 
certain cases, of which the 
following are examples, it 
is necessary to calculate 

the shear also, and combine it with the bending stress. 

For an I-beam the static moment / ydF is nearly as great directly 

under the flange as for a section through the neutral axis ; and there- 
fore, by formula (28), the shear is very large at this point, as shown 
on the shear diagram in Fig. 47. Hence the shear and bending 
stress are both large under the flange, and the resultant stress at 
this point may, in some cases, exceed that at the outer fiber. 

Again, if a beam is very short in comparison with its depth, or if 
the material of which it is made offers small resistance to shear in 
certain directions, as in the case of a wooden beam parallel to the 
grain, a special investigation of the shear must be made. For instance, 
consider a rectangular wooden beam of length /, breadth I, and depth h, 
bearing a single concentrated load P at its center. Then the total 



ANALYSIS OF STRESS IN BEAMS 



63 



shear on any section is > and the maximum bending moment is -- 
2 4 

Hence the maximum unit normal stress is 



3 PI 



^ _ 

P ~'~ I ' 2 ~ 2 



P C z bli* 

Also, since Q = and / ydF = > the maximum unit shear is 

2 Jo 8 

j. Cy*V=**. 

II J " 46fc 

Now let /c denote the ratio between the tensile strength in the direc- 
tion of the fiber and the shearing strength parallel to the fiber. 
Then, in order that the beam shall be equally safe against normal 
and shearing stress, p = tcq, or 



3PZ _ 3JP. 



whence 



FIG. 48 



20' 



In general, K is not greater than 10. If /c = 10, 1 = 5 h. Consequently, 
if the length of a beam is greater than 5 times its 

depth, the shear is not likely to cause rupture. ^~~ ^i ~J 

Problem 88. The bending moment and shear at a certain 
point in a Carnegie I-beam, No. B 2, of the dimensions 
given in Fig. 48, are M= 200,000 ft. Ib. and Q = 15,000 Ib. 
respectively. Calculate the maximum normal stress and the 
equivalent stress for a point directly under the flange, and 
compare these values with the normal stress in the extreme 
fiber. 

Solution. From the Carnegie handbook, the moment of 
inertia of this section about a neutral axis perpendicular 
to the web is I = 1466.5 in. 4 . Consequently, the normal 
stress in the extreme fiber is 

Me 2,400,000(10) 
I 1466.5 

and the normal stress at a point P under the flange is 

2,400,000(9.35) = 
1466.5 



64 STRENGTH OF MATERIALS 

A 

Neglecting the rounded corners, 



S*2 /* 

/ ydF= I 

Jk */9. 



Consequently, from formula (28), the unit shear at P is 



At the point P, therefore, p x = 15,300 Ib./in. 2 , p y = 0, and q = 64 Ib./in. 2 . 
Hence, from formula (7), Article 26, 

Pmax - y + \ V4^ + p2 = 1 5 , 30 4 Ib./in. 2 . 

To calculate the equivalent stress it is necessary to find the principal stresses, 
which are, from the above, 

p l = 15,304 lb./in. 2 and p 2 = - 2 lb./in. 2. 

Hence, from formula (14), Article 35, for ra = 3 the equivalent stress at P is 
p e = 15,305 lb./in. 2. 

58. Oblique loading. If, for any cross section, the plane of the 
external bending moment does not pass through a principal axis of 

the section, the loading is said 
to be oUique. In this case the 
bending moment M can be re- 
solved into components parallel 
to the principal axes, namely, 
M cos a and M sin #, where a 
is the angle which the plane 
containing M makes with one 
of the principal axes. 

For materials which conform to Hooke's law it has been found 
that the stress due to several sets of external forces can be calculated 
for each set separately and then combined into a single resultant. 
This is called the law of superposition. Applying this law to the 
present case, 

Jfcoso; M sin a Mcosa Msina 




FlG 49 



/OA\ 
(30) 






where e v , e z are the distances of the extreme fibers of the beam from 
the axes of Y and Z respectively, and S y , S z are the corresponding 
section moduli. 



ANALYSIS OF STRESS IN BEAMS 65 

Problem 89. In an inclined railway the angle of inclination with the horizontal 
is 30. The stringers are 10 ft. 6 in. apart, inside measurement, and the rails are 
placed 1 ft. inside the stringers. The ties are 8 in. deep and 6 in. wide, and the 
maximum load transmitted by each rail to one tie is 10 tons. Calculate the maxi- 
mum normal stress in the tie. 

Solution. The bending moment is the same for all points of the tie between the 
rails, and is 20,000 ft. Ib. From Problem 66, S z = 64 in. 3 and S y = 48 in. 8 . There- 
fore, from equation (30), 



240,000 (\ 240,000 (-\ 






- 5744 



59. Eccentric loading. If the external forces acting on any cross 
section reduce to a single force P, perpendicular to the plane of the 
section, but not passing through its center of gravity, this force is 
called an eccentric load. Let B denote the point of application of the 
eccentric load P, and let y'z 1 denote the coordinates of B. Then the 
eccentric force P acting at B can be replaced by an equal and parallel 
force acting at the center of gravity C of the section, and a moment 
whose plane is perpendicular to the-, section. This moment can then 
be resolved into two components parallel to the principal axes, of 
amounts Py' and Pz' respectively. Therefore, by the law of super- 
position, the intensity of the stress at any point (y, z) of the cross 

section is 

P Pz f Py' 

P = F + ^' Z + ^ y ' 
or, since I = Ftf, 



At the neutral axis the stfSs^s zero, and consequently 1 H - + ~- 

*y * 

must be zero ; or, since the semi-axes of the inertia ellipse are a = t y 
and b = t gt this condition becomes 

(3D P + g-'- . . 

This condition must be satisfied by every point on the neutral axis, 
and is therefore the equation of the neutral axis. To each pair of values 
of y' and z', that is, to each position of the point of application B of 
the eccentric load, there corresponds one and only one position of the 
neutral axis. 



66 



STRENGTH OF MATERIALS 



z 2 if 

If the point B lies on the ellipse + 7^ = 1, its coordinates must 

or lr 

satisfy this equation, and, consequently, 

(32) ^ + ^ = I- 

In this case the neutral axis passes through a point on the ellipse 
diametrically opposite to B ; for if z', y' are substituted for y 
and z in equation (31), it is evident that the condition (32) is satisfied. 

z 2 if 
The tangent to the ellipse -f- ^ = 1 at the point z 1 , y' is 

tyd ?/7/ 

+ ^- = 1, which is identical with equation (31). Consequently, 

if B lies on the inertia ellipse, the neutral axis corresponding to B is 
tangent to the ellipse at the point diametrically opposite to B. 

From equation (31), the slope of the tan- 
gent is found to be 



If, then, the point B moves out along a radius 
CB, z' and y' increase in the same ratio, and 
consequently the slope is constant ; that is to 
say, if B moves out along a radius, the neu- 
tral axis moves parallel to itself. 

As z r and y f increase, z and y must de- 
crease, for the products zz f and yy f must be 
constant in order to satisfy equation (31). 
Therefore the farther B is from the center of 
gravity, the nearer the corresponding neutral axis is to the center 
of gravity, and vice versa. 

If, in Fig. 50, TN is the neutral axis corresponding to B, it fol- 
lows, from the above, that CB CT is a constant wherever B is on the 
line BT. But if B lies on the ellipse, the corresponding neutral axis 
is tangent to the ellipse at the point diametrically opposite to B, and 
in this case the above product becomes CM 2 . Therefore 

(33) CB.CT 




FIG. 50 



From this relation, the position of the neutral axis can be determined 
when the position of the point B is given. 



ANALYSIS OF STKESS IN BEAMS 67 

60. Antipole and antipolar. The theorems in the preceding para- 
graph prove that if the point of application of an eccentric load lies 
outside, on, or within the inertia ellipse, the corresponding neutral 
axis cuts this ellipse, is tangent to it, or lies wholly outside it. 
This relation is analogous to that of poles and polars in analytical 
geometry, except that in the present case the point and its corre- 
sponding line lie on opposite sides of the center instead of on the 
same side. For this reason the point in the present case is called 
the antipole, and its corresponding line the antipolar. 

The following theorem is analogous to a well-known theorem of 
poles and polars. 

If the antipole moves along a fixed straight line, the antipolar 
revolves about a fixed point. Conversely, if the antipolar revolves 
about a fixed point, the antipole moves along a fixed straight line. 

If the antipole moves to infinity, the antipolar, or neutral axis, 
passes through the center of gravity of the section, which is the 
ordinary case of pure bending strain. The bending moment in this 
case can be considered as due to an infinitesimal force at an infinite 
distance from the center of gravity. 

If the antipole coincides with the center of gravity, the neutral 
axis lies at infinity, which means that the stress is uniformly dis- 
tributed over the cross section. 

Since the stresses on opposite sides of the neutral axis are of oppo- 
site sign, if the neutral axis cuts the cross section, stresses of both 
signs occur (i.e. both tension and compression), whereas if the neutral 
axis lies outside the cross section, the stress on the section is all of 
the same sign (i.e. either all tension or all compression). 

61. Core section. Let it be required to find all positions of the 
point of application of an eccentric load such that the stress on 
the cross section shall all be of the same sign. From the preceding 
article, the condition for this is that the neutral axis shall not cut 
the cross section. If, then, all possible lines are drawn touching the 
cross section or having one point in common with it, and the anti- 
poles of these lines are found, the locus of these antipoles will form 
a closed figure, called the core section. 

For a point within or on the boundary of the core section the neu- 
tral axis lies entirely without the cross section, or, at most, touches it, 



68 



STEENGTH OF MATERIALS 



and consequently stress of only one sign occurs. For a point without 
the core section the corresponding neutral axis cuts the cross section 
and it is subjected to stresses of both signs. 

Problem 90. Construct the core section for a rectangular cross section of breadth 
b and height h (Fig. 51). 

Solution. From Problem 56, I z = > I y = and the corresponding radii of 



12 



gyration are % = = andt^ = 
F 12 12 



ellipse are t g = 



h b 

and t v = ' 

2V3 



2\/3 



Consequently, the semi-axes of the inertia 
Having constructed the inertia ellipse, the 



vertices of the core section will be antipoles of the lines PQ, QR, RS, and SP. 

P Q 





From Article 59, the antipole of PQ is determined by the relation OA OE = Off 2 , or, 

since OE=- and OH=t e = -^ ,OA = -. Similarly, OC = - and OB = OD = -. 
2 2 V3 6 66 

Thus the core section is the rhombus A BCD, of which the vertices A, B, C, D are 
the antipoles of the lines PQ, PS, SR, QR respectively, and the sides AB, BC, CD, 
DA are the antipolars of the points P, S, R, Q respectively. 

Problem 91. Construct the core section for the T-shape in Problem 71. 

Solution. Six lines can be drawn which will have two or more points in com- 
mon with the perimeter of the T-shape without crossing it, namely, PQ, QR, RT, 
TU, US, and SP (Fig. 52). The vertices A, B, C, D, E of the core section are 
then the antipoles of these six lines respectively. 

Problem 92. Construct the core section of the I-beam in Problem 72. 

Problem 93. Construct the core section for the channel in Problem 73. 

Problem 94. Construct the inertia ellipse and core section for a circular cross 
section. 

62. Application to concrete and masonry construction. Since con- 
crete and masonry are designed to carry only compressive stresses, it 



ANALYSIS OF STRESS IN BEAMS 



69 



is essential that the point of application of the load shall lie within 
the core section. 

Consider a rectangular cross section of breadth ~b and height h. 
For the gravity axes MM and NN (Fig. 53) the corresponding mo- 
ments of inertia are 



I = 

m ~ 12 

Hence the radii of gyration are 
b 



Vl2 



= .28875 



A 

and 



and 






= ^ = .2887^, 
Vl2 




and the inertia ellipse is constructed on 
these as semi-axes. To determine the core 
section it is sufficient to find the antipole 
of each side of the cross section PQRS. 
Suppose A is the antipole of PQ, B the 
antipole of PS, etc. Then, by Article 60, 
the antipole of any line through P, such 
as LL, lies somewhere on AB ; that is to 
say, as the line PQ revolves around P to 
the position PS, its antipole moves along 
AB from A to B. The core section in the 
present case is thus found to be the rhom- 
bus ABCD. 

From Article 59, OC> OK = OT 2 = , since the semi-axes of the 

7 -L 

ellipse are the radii of gyration. But OK=-\ hence OC = - and 

7 7 2i D 

A C = - Similarly, BD = - - This proves the correctness of the 
o o 

rule ordinarily followed in masonry construction, namely, that in order 
to insure that the stress shall all be of the same sign, the center of 
pressure must fall within the middle third of the cross section. 

63. Calculation of pure bending strain by means of the core 
section. Let Fig. 54 represent the cross section of a beam subjected 
to pure bending strain. In this case the neutral axis passes through 
the center of gravity of a cross section, and therefore, from Article 60, 
the strain can be considered as due to an infinitesimal force at an 
infinite distance from the origin. Under this assumption the stress 



TO 



STRENGTH OF MATERIALS 



due to pure bending strain can be readily calculated by means of 
the core section, as follows. 

Suppose the external bending moment M lies in a plane perpen- 
dicular to the plane of the cross section and intersecting it in the 
line MM. Then, assuming that M is due to an infinitesimal force 
whose point of application is at an infinite distance from in the 
direction OM, the antipolar of this point will be the diameter of the 
inertia ellipse conjugate to MM. It is proved in analytical geometry 
that the tangent at the end of a diameter of a conic is parallel to the 
conjugate diameter. Therefore, if BT is tangent to the inertia ellipse 
at B, and NN is drawn through parallel to BT, NN will be the 

diameter conjugate to MM. Since the 
greatest stress occurs on the fiber most 
distant from the neutral axis, the maxi- 
mum stress will occur at P or E. Through 
P draw PA parallel to NN and intersect- 
ing JOf in A. Then, from Article 59, 



R 



s 




/N 




or, taking the projections of OA, OK, and 
OB on a line perpendicular to NN, 



FIG. 54 



e OKsma = (OB 

where e is the perpendicular distance of PA from 0. But OB sin a 
is the distance of the tangent BT from 0, and, by Article 49, this 
distance is the radius of gyration t corresponding to the axis NN. 
Therefore 



(34) 



e- OKsma = ^ = -2, 



where F is the area of the section and I n is its moment of inertia 
with respect to NN. The component of the external moment M per- 
pendicular to NN is M since. Hence, equating this to the internal 
moment, 

(35) Msma=^ Cy(ydF) = ^ Cy*dF = ^, 
e J e J e 

where p is the stress at the distance e from the neutral axis. Sub- 
stituting in equation (34) the value of I n obtained from equation (35), 



ANALYSIS OF STRESS IN BEAMS 71 

e- M siner 



whence 
(36) 

If, in the handbooks issued by iron and steel companies, the 
inertia ellipse and core section were drawn on each cross section 
tabulated, the calculation of the 
maximum bending stress by for- 



mula (36) would be extremely A B ^ ^ D 

simple, requiring merely the 
measurement of the distance OK. 

Problem 95. Calculate the maxi- 
mum bending stress in Problem 89 by 
means of the core section. 

Solution. The loading is as represented in Fig. 56, in which the portion BC 
is subjected to pure bending strain. From Problem 89, M = 20,000 ft. Ib. and 
F = 48 in. 2 . From the diagram of the core section drawn to scale, OK is found 
to measure .9 in. Therefore, from formula (36), p Q = 6555 lb./in. 2 . 

64. Stress trajectories. In Article 27 the principal stresses at any 
point in a body were denned as the maximum and minimum normal 
stresses at this point. Lines which everywhere have the direction 
of the principal stresses are called stress trajectories. 

In order to determine the stress trajectories, a number of cross 
sections of the body are taken, and the shear and normal stress cal- 
culated for a number of points in each section. The directions which 
the principal stresses at these points make with the axis of the body 
can then be found by formula (6), Article 26, as explained in Prob- 
lem 39. The stress trajectories are thus determined as the envelopes 
of these tangents. 

Since the principal stresses at any point are always at right angles, 
the stress trajectories constitute a family of orthogonal curves. 

65. Materials which do not conform to Hooke's law. The preced- 
ing articles of this chapter are based on Hooke's law, and consequently 
the results are applicable only to materials which conform to this 
law, such as steel, wrought iron, and wood. Other materials, such as 
cast iron, stone, brick, cement, and concrete, are so lacking in homo- 
geneity that their physical properties are very uncertain, differing not 



72 STRENGTH OF MATERIALS 

only for different specimens of the material but also for different por- 
tions of the same specimen. For this reason it is impossible to apply 
to such materials a general method of analysis with any assurance 
that the results will approximate the actual behavior of the material. 
For practical purposes, however, the best method is to calculate the 
strength of such materials by the formulas deduced above, and then 
modify the result by a factor of safety so large as to include all 
probable exceptions. 

The behavior of cast iron is more uncertain than that of any other 
material of construction, and it must therefore be used with a larger 
factor of safety. If two pieces from the same specimen are subjected 
to tensile strain and to cross-bending strain respectively, it will be 
found that the ultimate strength deduced from the cross-bending 
test is about twice as great as that deduced from the tensile test. 
The reason for this is that the neutral axis does not pass through 
the center of gravity of a cross section, lying nearer the compression 
than the tension side, and also because the stresses increase more 
slowly than their distances from the neutral axis. If, then, it becomes 
necessary to design a cast-iron beam, the ultimate tensile strength 
used in the calculation should be that deduced from bending tests. 

For materials such as concrete, stone, and cement, the most 
rational method of procedure is to introduce a correction coefficient 

77 in formula (18) and put 

Me 

p = r\ 

where it has been found by experiment that for granite 77 = .96, for 
sandstone rj = .84, and for concrete 77 = .97.* 

66. Design of reenforced concrete beams. Since concrete is a mate- 
rial which does not conform to Hooke's law, and moreover does not 
obey the same elastic law for tension as for compression, the exact 
analysis of stress in a plain or reenforced concrete beam would be 
much more complicated than that obtained under the assumptions of 
the common theory of flexure. The physical properties of concrete, 
however, depend so largely on the quality of material and workman- 
ship, that for practical purposes the conditions do not warrant a rig- 
orous analysis. The following simple formulas, although based on 

* Foppl, Festigkeitslehre, p. 144. 



ANALYSIS OF STRESS IN BEAMS 



73 



approximate assumptions, give results which agree closely with exper- 
iment and practice. 

Consider first a plain concrete beam, that is, without reenforcement. 
The elastic law for tension is in this case (see Fig. 56) 



and for compression 





To simplify the solution, however, assume 

the straight-line law of distribution of 

stress, that is, assume m x = m 2 1. Note, 

however, that this does not make the 

moduli equal. Assume also that cross sections which were plane 

before flexure remain plane after flexure (Bernoulli's assumption), 

which leads to the relation 



where e c and e t denote the distances of the extreme fibers from the 
neutral axis (Fig. 56). 

Now let the ratio of the two moduli be denoted by n, that is, let 



A. 



Then 



Pt 



For a section of unit width the resultant compressive stress R c on 
the section is R c = \ p c e c , and similarly the resultant tensile stress 
R t is E t = ^p t e t . Also, since R c and R t form a couple, R c = R t . Hence 

*Y) > *Y) 

p c e c = p t e ( , or *- = -*; and equating this to the value of the ratio - 

Pt e c Pt 

obtained above, we have 

e.= e.'vn. 



74 STRENGTH OF MATERIALS 

Since the total depth of the beam h is h = e c + e t) we have, therefore, 

e c = h e c ^n, whence 

h 

e..= 



and, similarly, e t =h - -j=> whence 



Now, by equating the external moment M to the moment of the 
stress couple, we have 



whence, by solving for the unit stresses p c and p t , 



e 

ii 

or, solving one of these two relations for A, say the first, we have 



A to 1+ v^). 

M r>~ 



For ordinary concrete n may be taken as 25. Also, using a factor 
of safety of 8, the working stress p c becomes p c = 300 lb./in. 2 Substi- 
tuting these numerical values in the above, the formula for the depth 
of the beam in terms of the external moment takes the simple form 

VM 



h being expressed in inches, and M in inch pounds per inch of width 
of beam. 

Problem 96. A plain concrete slab, supported on two sides only, has a 12-ft. 
span and carries a load of 200 lb./ft. 2 Find the required thickness. 

Solution. The load is fff lb./in. 2 , and hence for a strip 1 in. wide, the maxi- 
mum moment is M = = 3600 in. Ib. Consequently the required depth h is 

h = = 15 in. 
4 




ANALYSIS OF STRESS IN BEAMS 75 

For a reenforced concrete beam the tensile strength of the concrete 
may be neglected. Let E c and E s denote the moduli of elasticity for 
concrete and steel respectively, and let 

. Rf. 

-^- n. Then if x denotes the distance 
E c 

of the neutral axis from the top fiber 

(Fig. 57), the assumptions in this case F 

are expressed by the relations 

S X Pc J? A Ps _ J? 

s s h x ' s c s s ~ 

whence 

s c p c E s p c x 

= - - = -w. - = 

or, solving for x. 



Now if F denotes the area of steel reinforcement per unit width of 
beam, then 

R s = p s F and R c =\p c x-, 

and consequently, since U c = JR S , 



Moreover, equating the external moment M to the moment of the 
stress couple, we have 



Substituting the value of x in either one of these expressions, say 
the first, we have 

1 , np c /, h np c \ . 
M=-p c h - *-s ( A - - ~ 
2^" p,+ np c \ 3p, + n 

whence, solving for h, 



== p s +np c 

>w 



76 STRENGTH OF MATERIALS 

For practical work assume n = 15, p c = 500 lb./in. 2 (factor of safety 
of 5), and p s = 15,000 lb./in. 2 (factor of safety of 4). Substituting these 
numerical values in the above, the results take the simple form 



where H denotes the total depth of the beam in inches, d is the diam- 
eter of the reenf orcement in inches, and M is the external moment in 
inch pounds per inch of width. 

In designing beams by these formulas first find h, then F, and 
finally H. 

Problem 97. A reenforced concrete slab, supported on two sides only, has a 
12-ft. span and carries a load of 200 lb./ft. 2 Find the required thickness of slab 
and area of metal reinforcement per foot of width. 

Solution. As in the preceding example, the maximum moment is M = 3600 in. Ib. 

Consequently, h= .116 V^7 = 6.96 in.; also F = '' in. 2 per inch of width, or 

in. 2 per foot of widtli = .464 in. 2 /ft.; and hence the diameter of the reen- 
180 

f orcement is d = f in. for round rods spaced one foot apart. Finally, the total 
depth of slab is II 6.96 + f + = 7.84 in., say 8 in. 

An interesting application of these formulas is the comparison of 
the calculated position of the neutral axis in a reenforced concrete 
beam with that determined experimentally. It has been shown by 
experiment that when a reenforced concrete beam is loaded, minute 
cracks appear extending upward from the bottom, showing that prac- 
tically all the tensile stress is carried by the reinforcement. To 
render this more obvious, before the concrete is put in, place one or 
more sheets of pasteboard vertically in the mold in which the beam 
is made, extending completely across the mold and upward from the 
bottom to within a distance of the top at least equal to the value of 
x given by the above formulas. This eliminates entirely the tensile 
strength of the concrete, which is the assumption upon which the 
above formulas are based; and when the beam is loaded the exten- 
sion of the reenforcetrient causes a crack to appear plainly along the 



ANALYSIS OF STRESS IN BEAMS 



77 



pasteboard. Since this crack 
must end at the neutral axis, 
the position of this axis is thus 
approximately determined ex- 
perimentally and maybe used to 
verify the calculated value of x. 

EXERCISES ON CHAPTER III 

Problem 98. A structural steel- 
built beam is 20 ft. long and has the 
cross section shown in Fig. 58. Com- 
pute its moment of resistance and 
find the safe uniform load it can 
carry per linear foot for a factor of 
safety of 5. 

Problem 99. The cast-iron bracket shown in Fig. 59 has at the dangerous section 
the dimensions shown in the figure; Find the maximum concentrated load it can 
carry with a factor of safety of 15. 



1 




r: 






n^-f t 
&" 




l" 


[ 

i" 12' 

1 j 






C | 









FIG. 58 



131/2-- 




FIG. 59 



78 



STRENGTH OF MATERIALS 



Problem 100. Find the proper dimensions for a wrought-iron crank shaft of 
dimensions shown in Fig. 60 for a crank thrust of 1500 Ib. and a factor of 
safety of 6. 

Problem 101. A wrought-iron pipe 1 in. in ex- 
ternal diameter and T ^ in. thick projects 6 ft. from 
a wall. Find the maximum load it can support at 
the outer end. 

Problem 102. The yoke of an hydraulic press 
used for forcing gears on shafts is of the form and 
dimensions shown in Fig. 61. The yoke is horizon- 
tal with groove up, so that the shaft to be fitted lies 
in the groove, as shown in plan in the figure. The 
ram is 32 in. in diameter and under a water pressure 

of 250 lb./in. 2 Find the dangerous section of the yoke and the maximum stress 
at this section. 

Problem 103. Design a concrete conduit, 7 ft. square inside, to support a con- 
centrated load of 1000 Ib. per linear foot, and determine the size and spacing of 
the reinforcement. 

' Problem 104. A 10-in. I-bar weighing 40 Ib./f t. is supported on two trestles 15 ft. 
apart. A chain block carrying a 1-ton load hangs at the center of the beam. Find 
the factor of safety. 




FIG. 60 




PLAN 



END ELEVATION 



FIG. 61 



Problem 105. The hydraulic punch shown in Fig. 62 is designed to punch 
a f-in. hole in a f-in. plate. The dimensions of the dangerous section AB are as 
given in the figure. Find the maximum stress at this section. 

Problem 106. The load on a car truck is 8 tons, equally distributed between 
the two wheels (Fig. 63). The axle is of cast steel. Find its diameter for a factor 
of safety of 15. 

Problem 107. The floor of an ordinary dwelling is assumed to carry a load of 
50 lb./ft. 2 and is supported by wooden joists 2 in. by 10 in. in section, spaced 16 in. 
apart on centers. Find the greatest allowable span for a factor of safety of 10. 



ANALYSIS OF STRESS IN BEAMS 



79 




FIG. 62 



Problem 108. A wooden girder supporting the bearing partitions in a dwelling 
is made up of four 2-in. by 10-in. joists set on edge and spiked together. Find the 
size of a steel I-beam of equal strength. 

Problem 109. A factory floor is assumed to carry a load of 200 lb./ft. 2 and is 
supported by steel I-beams of 16 ft. span and spaced 4 ft. apart on centers. What 
size I-beam is required for a 
factor of safety of 4 ? 

Problem 110. Find the re- 
quired size of a square wooden 
beam of 14 ft. span to carry an 
axial tension of 2 tons and a 
uniform load of 100 lb./ft. 

Problem 111. A reenforced 
concrete beam 10 in. wide and 
22 in. deep has four 1^-in. 
round bars with centers 2 in. 
above the lower face. The 
span is 16 ft. The beam is 
simply supported at the ends. 
Find the safe load per linear 

foot for a working stress in the concrete of 500 lb./in. 2 , and also find the tensile 
stress in the reenforcement. 

Problem 112. A reenforced concrete flopr is to carry a load of 200 lb./ft. 2 over 
a span of 14 ft. Find the required thickness of the slab and area of the reenforce- 
ment for working stresses of 500 lb./in. 2 in the concrete and 15,000 lb./in. 2 in the 
reenforcement. 

Problem 113. A reenforced concrete beam of 16 ft. span is 18 in. deep, 9 in. 
wide, and has to support a uniform load of 1000 Ib. per linear foot. Determine the 
amount of steel reenforcement required, bars to have centers 2 in. above lower face 

of beam. 

Problem 114. Find the maximum mo- 
ment and maximum shear, and sketch the 
shear and moment diagrams for a canti- 
lever beam 8 ft. long, weighing 20 lb./ft., 
with concentrated loads of 200 and 300 
Ib. at 3 and 5 ft. respectively from the 
free end. 

Problem 115. Find the maximum mo- 
ment and maximum shear, and sketch the 
shear and moment diagrams for a canti- 
lever beam 12 ft. long, carrying a total 
uniform load of 50 lb./ft. and concen- 
trated loads of 200, 150, and 400 Ib. at 
distances of 2, 4, and 7 ft. respectively from the fixed end. 

Problem 116. A beam 30 ft. long carries concentrated loads of 1 ton at the left 
end, 1.5 tons at the center, and 2 tons at the right end, and rests on two supports, 
one 4 ft. from the left end and the other 6 ft. from the right end. Sketch the shear 
and moment diagrams and find the maximum shear and maximum moment. 




FIG. 63 



80 STRENGTH OF MATERIALS 

Problem 117. A beam 20 ft. long bears a uniform load of 100 Ib. per linear foot 
and rests on two supports 10 ft. apart and 5 ft. from the ends of the beam. Find 
the maximum moment and shear, and sketch the shear and moment diagrams. 

Problem 118. Find the maximum moment and maximum shear, and sketch the 
shear and moment diagrams for a simple beam 10 ft. long, bearing a total uniform 
load of 100 Ib. per linear foot and concentrated loads of 1 ton at 4 ft. from the left 
end and 2 tons at 3 ft. from the right end. 



NOTE ON SHEAR AND MOMENT DIAGRAMS 

It is important to be able to sketch readily by inspection the shear and moment 
diagrams corresponding to any given loading. To acquire this ability it is only 
necessary to observe the characteristic features of such diagrams. The more im- 
portant of these are as follows : 

The slope of the moment curve is equal to the shear. From this, the following 
conclusions are obtainable. 

Where the moment is a maximum the shear is zero. Note, however, that for 
concentrated loads the moment has no calculus maximum. In this case, where the 
moment has its greatest value, the shear passes through zero because the slope of 
the moment diagram necessarily changes from positive to negative at this point. 

Where the moment is constant the shear is zero. 

For a uniform load the moment diagram is a parabola and the shear diagram 
is an inclined line whose slope is equal to the load per unit of length. Mathemat- 
ically this means that the parabola is a curve whose slope changes uniformly from 
point to point. 

For concentrated loads the moment diagram is a broken straight line, and the 
shear diagram is a series of horizontal lines or steps. 

For uniform and concentrated loads combined, the moment diagram is a series 
of parabolic arcs, and the shear diagram is a series of inclined lines or sloping steps. 

At the ends of a simple beam the moment is always zero. 

Where the moment diagram crosses the axis, the elastic curve or center line of 
the beam has a point of inflection ; that is to say, the beam is curved upward on 
one side of this point and curved downward on the other side. Such a point is 
called a point of contrqflexure. The tensile stress changes from the bottom to the 
top on opposite sides of a point of contraflexure, and such points are therefore 
of especial importance in the case of reenforced concrete beams, as the reenforce- 
ment must always follow the tensile stress. 

The area subtended by the shear diagram up to any point is equal to the mo- 
ment at this point, since = Q and therefore M = C Odx. 
dx J 



CHAPTER IV 

FLEXURE OF BEAMS 

67. Elastic curve. If a beam is subjected to transverse loading, its 
axis is bent into a curve called the elastic curve. The differential equa- 
tion of the elastic curve is 
found as follows. 

Let ABDE (Fig. 64) rep- 
resent a portion of a bent 
beam limited by two adja- 
cent cross sections AB and 
DE, and let C be a point 
in the intersection of these 
two cross sections. Then 
C is the center of curva- 
ture of the elastic curve 
FH. Let dp denote the 
angle ACE y and through 
H draw LK parallel to A C\ 
then the angle LHE is also equal to d(3. Since the normal stress is 
zero at the neutral axis, the fiber FH is unchanged in length by the 
strain. Therefore, from Fig. 64, the change in length of a fiber at a 
distance y from the elastic curve is yd{3, where dp is expressed in 
circular measure. Consequently, the unit deformation of such a fiber is 



s = 




dx 



By Hooke's law, = E where p = - ; hence 
s J. 



Is 



81 



82 



STRENGTH OF MATERIALS 



Inserting in this expression the value of s just found, ^ x = E ; 
whence 



Let the radius of curvature CF of the elastic curve be denoted by p. 
Then pd@ = dx, and inserting this value of dfi in the above equation, 
it becomes 



From the differential calculus, the radius of curvature of any curve 
can be expressed by the formula 



But since the deformation of the beam is assumed to be small, the 
slope of the tangent at any point of the elastic curve is small ; that 

is to say, -j- is infinitesimal, and consequently (-j-\ can be neglected 

in comparison with ^ Under this assumption p = , and there- 
m- i dx d y 



, El 1 , 

fore = p = - ; whence 
M d? 



dx? 



(37) 



dx 



= M, 



which is the required 
differential equation of 
the elastic curve. 

In what follows the 
external bending mo- 
ment M is assumed to 
be negative if it tends 

Flo 65 to revolve the portion 

of the beam under 

consideration in a clockwise direction, and positive if the revolution 

is counter-clockwise. 




FLEXUKE OF BEAMS 



83 



Problem 119. Find the equation of the elastic curve and the deflection at the 
center of a simple beam of length Z, bearing a single concentrated load P at its 
center. 

Solution. The elastic curve in this case consists of two branches, AB and BC 
(Fig. 65). 

Consider the portion of the beam on the left of any section mn, distant x from 

P 
the left support. Then M = R^x = x, and consequently the differential 

equation of the branch AB of the elastic curve is 



Integrating twice, 



dx 2 
dy 



and 



C 2 . 



At .B, x = - and = 0, since the tangent at B is horizontal. Substituting these 



dx 



PI* 



values in the first integral, Ci = - At A, x = and y = ; hence C 2 = 0. Con- 
sequently, the equation of the left half of the elastic curve is 

Px . 



y = 



48 El 



The deflection D at the center is the value of y f or x = - ; hence 



48 El 

Problem 120. Find the 
equation of the elastic 
curve and the maximum 
deflection for a cantilever 
of length Z, bearing a sin- 
gle concentrated load P at 
the end. 

Problem 121. Find the 
equation of the elastic curve 
and the maximum deflec- -^ IG - 6 ^ 

tion for a simple beam of 
length Z, bearing a single concentrated load P at a distance d from the left support. 

Solution. The elastic curve in this case consists of two branches, AB and BC 




(Fig. 66). For a point in AB distant x from the left support, M = 
Therefore 

P(l-d)x 

JTjJ. = 

dx2 
Integrating twice, 



-P(l-d)x 
I 



84 STRENGTH OF MATERIALS 

and Ely = h CiX + C 2 - 

6 1 

At A, x = and y = ; therefore C 2 = 0. In order to determine Ci it is necessary 
to find the equation of BC. 

Taking a section on the right of 2?, M = - - , and consequently 



_ Pd(l-x) 

M> ~ 

Integrating twice, 



Pd /Zr 2 rr s \ 
and Ely = - I - ~\ + C 3 x + C 4 . 

At C, x = I and y = ; therefore C 4 = C 3 Z. 

3 

Now at B both branches of the elastic curve have the same ordinate and the 
same slope. Therefore, putting x = d in the above integrals and equating the slopes 
and ordinates of the two branches, 

-P(l-d)d* _ _ I 

~ 21 1 ~~~ 



Gl 
Solving these two equations simultaneously for C\ and 

61 



Substituting these values of C\ and <7 3 in the above integrals, the equation of the 
branch AB becomes, after reduction, 



and the equation of BC becomes 



Since the load is not at the center of the fleam, the maximum deflection will 
occur in the longer segment. Moreover, at the point of maximum deflection the 

tangent is horizontal, that is, = 0. Therefore, equating to zero the first differ- 
ential coefficient of the branch AB, 



FLEXUKE OF BEAMS 



85 



from which the distance of the point of maximum deflection from the left support 
is found to be 



and the deflection at this point is 

D = 



ami 



Problem 122. Find the equation of the elastic curve and the maximum deflec- 
tion for a simple beam of length /, bearing a uniform load of w Ib. per unit 
of length. 

Problem 123. Find the equation of the elastic curve and the maximum deflec- 
tion for a cantilever of length I, uniformly loaded with a load of w Ib. per unit 
of length. 

Problem 124. A Carnegie I-beam, No. B 13, is 10 ft. long and bears a load of 
25 tons at its center. Find the deflection of the point of application of the load. 

NOTE. From the Carnegie handbook, the moment of inertia of the beam about a 
neutral axis perpendicular to the web is I = 84.9 in. 4 

Problem 125. Find the deflection of the beam in the preceding problem at a 
point 4 ft. from one end. 

68. Limitation to Bernoulli's assumption. In Article 39 it was 
stated that Bernoulli's assumption formed the basis of the common 




FIG. 68 

theory of flexure. In the case of a prismatic beam subjected to pure 
bending strain, this assumption is rigorously correct. For if the oppo- 
site faces of a prism ABCD (Fig. 67) are acted upon by equal bend- 
ing moments of opposite sign, both faces must, by reason of symmetry, 
remain plane and take a position such as A'B'C'D' in the figure. 

However, if shearing stress also occurs, Bernoulli's assumption is 
no longer absolutely correct. In Article 55 it was proved that the 
distribution 'of shear over any cross section limited by parallel sides 
varies as the ordinates to a parabola. Consequently, if the beam is 
supposed cut into thin layers by horizontal planes, as represented in 



86 STRENGTH OF MATERIALS 






Fig. 68, the shear will tend to slide these layers one upon another. 

By Hooke's law, the amount of this sliding for different layers will 

also vary as the ordinates to a parabola, 
being zero at top and bottom and a maxi- 
mum at the center. Therefore, if the 
elongations and contractions of the fibers 
due to bending stress are combined with 
the sliding due to shear, the resultant 
deformation of the prism will be as rep- 
resented in Fig. 69. 

69. Effect of shear on the elastic curve. In addition to the hori- 
zontal shearing stress acting at any point in a beam, there is a shear- 
ing stress of equal intensity acting in a vertical direction. The effect 
of this vertical shear is to slide each cross section past its adjacent 
cross section, as represented in Fig. 70, and 

thus increase the deflection of the beam. 

In Article 83 a general formula is derived 
by means of which the shearing deflection 
can be calculated in any given case. It is 
found, however, that in all ordinary cases the 
shearing deflection is so small that it can be D 
neglected, in comparison with the deflection 
due to bending strain. The point to be re- 
membered, then, is that the shearing deflec- FlG 70 
tion is negligible but not zero. 

In precise laboratory experiments for the determination of Young's 
modulus it should always be ascertained whether or not the shearing de- 
formation can be neglected without affecting the precision of the result. 

70. Built-in beams. If the ends of a beam are secured in such a 
way as to be immovable, the beam is said to be built-in. Examples of 
built-in beams are found in reenforced concrete construction, in which 
all parts are monolithic. Thus a floor beam in a building constructed 
of reenforced concrete is of one piece with its supporting girders, and 
consequently its ends are immovable. 

Since the tangents at the ends of a built-in beam are horizontal, 

dit 

-j- at these points. Also, from Fig. 71, it is obvious that the 



FLEXURE OF BEAMS 



87 



elastic curve of a built-in beam differs from that for a simple beam in 
having two points of inflection, A and B. At these points the curvature 

d\ 

is zero, that is, - = 0, 
dx 2 

and consequently the bend- 
ing moment is also zero, 

since EI^- = M. 

dx 2 FIG. 71 




Problem 126. Find the equation of the elastic curve and the maximum deflec- 
tion for a beam of length Z, fixed at both ends and bearing a uniform load of w Ib. 
per unit of length. 

Solution. Let M a and M b denote the moments at the supports (Fig. 72). The 
vertical reactions at the supports are each equal to 

Consequently, the bending moment at a point distant x from the left support is 

_ wlx wx 2 

x a ~~^~ ~2~ 



and therefore 



wlx wx* 




FIG. 72 



Integrating, 



dx 



4 



6 



At A, x = and = ; therefore Ci = 0. At B, x = I and = ; therefore 

wl 2 dx dx 

M a = Substituting this value of M a in the above integral, and integrating again, 



. 

At A, x = and y = ; therefore C 2 = 0. Consequently, the equation of the elastic 
curve is, after reduction, 

_ wx 2 (I - x) 2 



88 STRENGTH OF MATERIALS 

Putting x = - in this equation, the maximum deflection is found to be 



384 El 

At the points of inflection ^ = 0. Therefore 
ax' 2 

wlx wx 2 
a ~ ~2 2~' 
whence 

x = - -L= = .2121 or .788J, 

which are the distances of the two points of inflection from the left support. 

Problem 127. A beam of length I is fixed at both ends and bears a single con- 
centrated load P at a distance d from the left end. Find the deflection at the 
point of application of the load. 

Problem 128. From the result of Problem 127, find the deflection at the point 
of application of the load when the load is at the center. 

Problem 129. A concrete girder 16 ft. long, 18 in. deep, and 12 in. wide is 
reenforced by two 1-in. twisted square steel rods near its lower face, and bears 
a uniform load of 250 Ib. per linear inch. The moment of inertia of the equiv- 
alent homogeneous section about its neutral axis (Article 49) is found to be 
I c = 7230 in. 4 Find the maximum deflection. 

71. Continuous beams. A continuous beam is one which is sup- 
ported at several points of its length, and thus extends continuously 
over several openings. If the reactions of the several supports were 
known, the distribution of stress in the beam and the equation of 
the elastic curve could be found by the methods employed in the 
preceding articles. The first step, therefore, is to determine the 
unknown reactions. General methods for determining these will be 
explained in Articles 72, 78, 80, and 81. The two following prob- 
lems illustrate special methods of treating the two simple cases 
considered. 

Problem 130. A beam i simply supported at its center and ends, and bears 
a single concentrated load P at the center of each span. Assuming that the 
supports are at the same level, find their reactions and the equation of the 
elastic curve. 

Solution. Let each span be of length Z, and assume the origin of coordinates 
at O (Fig. 73). Consider the portion of the beam on the right of a section mn, 

distant x from 0. Then, if x < - , 



FLEXUEE OF BEAMS 



89 



Integrating twice, 
(38) 

(39) 



At 0, x = and = ; therefore Ci = 0. Also at 0, x = and y = ; therefore 

Let x be greater than -. Then the differential equation of the branch AB 
becomes 

=-*<'-*> 

Integrating, 
(40) 



D 




FIG. 73 

At A both branches, OA and J.J5, have the same slope. Therefore, putting x = - 
in (38) and (40), and equating the values of thus obtained, 



whence 



Substituting this value of <7 3 in equation (40), and integrating again, 



At J. both curves have the same ordinate. Therefore, putting x = - in equations 
(39) and (41), and equating the values of y thus obtained, 



90 

whence 



STRENGTH OF MATERIALS 



PI* 



The equations of both branches of the elastic curve are now determined except 
that the reaction E 3 is still unknown. Since B is assumed to be on the same level 
with O, its ordinate is zero. Therefore, to determine JJ 3 , put x = I and y = in 
equation (41) ; whence 



From symmetry E\ = 3. Therefore 



Problem 131. Determine the reactions of the supports for a beam simply 
supported at its center and ends, and bearing a 'uniform load of w Ib. per unit 
of length. 

Solution. If the end supports were removed, the beam would consist of two 
cantilevers, AB and BC (Fig. 74), each of length I and bearing a uniform load. 




FIG. 74 



wl* 



From Problem 123, the deflection at the end of such a beam is D = -- But the 

reaction E 3 (or EI) must be of such amount as to counteract this deflection ; and, 
from Problem 120, the deflection at the end of a cantilever bearing a single concen- 



trated load J? 3 is D = 



whence 






Therefore 



From symmetry, EI = E s . Consequently, 

E 2 = 2wl- (Ei + Eg) = f wl 

Having found the reactions of the supports, the equations of the elastic curves can 
be determined as in the preceding problems. 

72. Theorem of three moments. The theorem of three moments 
is an algebraic relation between the bending moments at three con- 
secutive piers of a continuous beam. The theorem is due to Clapeyron, 



FLEXUKE OF BEAMS 



91 



and first appeared in the Comptes Rendus for December, 1857. The 
following is a simplified proof of the theorem for the case of 
uniform loading. 

Let A, B, C be three consecutive piers of a continuous beam at 
the same height, and let M a , M b) M c and R u , R b , R c denote the bend- 
ing moments 
and reactions 
at these three 
points respec- 
tively (Fig. 75). 
Also let l lt 1 2 de- 
note the lengths QA 
of the two spans 
considered, w lt w 2 

the unit loads on U - x - *j 

them, and Q' a , Q" 

the shears on the 

left and right of 

R a respectively, with a similar notation for the other supports. Then, 

taking A as origin, the differential equation of AB is 



(42) 
Integrating twice, 

(43) 

and 




FIG. 75 



-r 2 7> 3 on r* 

Ely = Jf. J + Qf- - ^J + Of + 



At A, x and y = ; hence (7 2 =0. At B, x = l^ and y = ; hence 



In equation (42), if x = l lt EI^ = M b . Therefore 



(44) 



Cf't'O 



If -) denotes the slope of the elastic curve AB at B, then, from 
equation (43), 



92 STRENGTH OF MATERIALS 



Similarly, by taking the origin at C and reckoning backward toward 
B, it will be found that 

(46) j^Jt + g^-SS, 

and 



Equating the values of ( -j- ] from equations (45) and (47), and elimi- 
\dxj b 

nating Q' and Q' e from the resulting equation by means of equations 
(44) and (46), 



whence 



which is the required theorem of three moments. 

If the beam extends over n supports, this theorem furnishes n 2 
equations between the n moments at the supports, the remaining two 
equations necessary for solution being furnished by the terminal con- 
ditions at the ends of the beam. 

Problem 132. A continuous beam of two equal spans bears a uniform load 
extending continuously over both spans. Find the bending moments and reac- 
tions at the supports. 

Solution. In the present case w\ = w> 2 = to, li = ^ = Z, and M a = M c = 0. Con- 
sequently, the theorem reduces to 



whence 

From equation (44), 



ryn . 

= (/ I -- J 

a 



8 a 2 

whence 



FLEXURE OF BEAMS 93 

From symmetry, R a = JR C , and consequently 

-R& = | wl. 

Problem 133. A continuous beam of four equal spans is uniformly loaded. Find 
the bending moments and reactions at the supports. 

Solution. The system of simultaneous equations to be solved in this case is 

M l = M 5 = 0, 



the solution of which gives 



Qi" = i i wrf, Qa' = \ I "I* Q 2 " 
JB, = fl 5 = Q t " = |i wl, E 2 = E 4 = Q 2 ' + Q 2 - = f /, B 8 = Q/ + Q 8 " = i| w i. 

Problem 134. A continuous beam of five equal spans is uniformly loaded. Find 
the moments and reactions at the supports. 

73. Work of deformation. In changing the shape of a body the 
points of application of the external forces necessarily move, and 
therefore do a certain amount of work called 
the work of deformation. 

To find the amount of this work of defor- 
mation for a prismatic beam, consider two adja- 
cent cross sections of the beam at a distance 
dx apart (Fig. 76). Suppose one of these cross 
sections remains stationary and the other turns 
through an angle d/3 with reference to the first. 
Then the change hi length of a fiber at a dis- 
tance y from the neutral axis is ydft, and therefore, by Hooke's law, 




_ p 
dx E 



where p is the intensity of the stress on the fiber. By the straight- 

line law, p = - > and hence 

Mdx 



Since one of the cross sections is assumed to be stationary, the stress 
acting on it does no work. On the other cross section the normal 



94 STEEXGTH OF MATERIALS 

stress forms a moment equal to M. This moment is zero when first 
applied, and gradually increases to its full value, its average value 
being \M. Therefore the work done by the normal stress on this 
cross section is 2 , 



Hence the total work of deformation for the entire beam is 



Problem 135. As an application of the above, find the deflection at the center 
of a simple beam of length /, bearing a single concentrated load P at the center. 

Solution. Let D denote the deflection at the center. Then the external work 
of deformation is 

W= | PD. 

Px 

At a point distant x from the left support the bending moment is M = , and 

consequently the internal work of deformation is 



96 El 



1 P 2 / 3 Pl s 

Therefore - PI) = ; whence 1) = 



2 96 El 48 El 

Problem 136. Find the internal work of deformation for a rectangular wooden 
beam 10 ft. long, 10 in. deep, and 8 in. wide, which bears a uniform load of 250 Ib. 
per foot of length. 

74. Impact and resilience. If the stress lies within the elastic 
limit of the material, the body returns to its original shape upon 
removal of the external forces, and the internal work of deformation 
is given out again in the form of mechanical energy. The internal 
work of deformation is thus a form of potential energy, and from 
this point of view is called resilience. The work done in straining a 
unit volume of a material to the elastic limit is called the modulus 
of elastic resilience of the material. 

It is therefore represented by the area under the strain curve up 
to the elastic limit, or, expressed as a formula, 

(stress at elastic limit) 2 

Mod. elas. resilience = - 

2 modulus of elasticity 



FLEXURE OF BEAMS 95 

When a load is suddenly applied to a beam, as when a body falls 
on the beam, or in the case of a railway train passing quickly over a 
girder, the deflection of the beam is much greater than it would be if 
the load was applied gradually, for in this case the full amount of 
the load is applied at the start instead of gradually increasing from 
zero up to this amount. Since the load is not sufficiently great to 
cause the beam to retain this deflection, the resilience of the beam 
causes it to vibrate back and forth until the effect of the shock dies 
away. The sudden application of a load is called impact, and the 
study of its effect is of especial importance in designing machines, 
railway bridges, or any construction liable to shocks. 

If a simple beam deflects an amount D under a load P suddenly 
applied, the work of deformation is PD. If the beam deflects the 
same amount under a load P' gradually applied, the work of defor- 

mation is i- P'D. Hence 

P' = 2 P. 

In other words, the strain produced in a beam by a load applied sud- 
denly is equivalent to the strain produced by a load twice as great 
applied gradually. In practical work P' is assumed to be about f P 
instead of 2 P, for it is impossible to apply a load instantaneously at 
the most dangerous section. 

If a body of weight P falls on a beam from a height h and pro- 
duces a deflection D, the work done by P is P(h + D}. Therefore, 
if P' is the amount of a static load which would produce the same 
deflection, 



In order to find P f from this equation D must be expressed in terms 
of P' and its value substituted in the above expression before solving 
for P 1 . 

Problem 137. A Cambria steel I-beam, No. B 33, is 12 ft. long and 10 in. deep, 
and has a moment of inertia about an axis perpendicular to the web of 122.1 in. 4 . 
What is the maximum load that can fall on the center of the beam from a height 
of 6 in. without producing a stress greater than 25,000 lb./in. 2 , if 75 per cent of the 
kinetic energy of the falling body is transformed into work of deformation ? 

Solution. Let P denote the weight of the falling body and P' the amount of a 
static load which would produce the same work of deformation. Then, since the 

TO/? -\fp P'le 4?)Z 

moment at the center of the beam is Jf = , p = - = - , whence P' = 



96 



STRENGTH OF MATERIALS 



The deflection of a beam bearing a static load P' at the center is 1) = 
(Problem 119), or, substituting in this the value of P', D = 

I /'. ' 



P'V 
48 El 

Assuming 



E = 30,000,000 lb./in. 2 , and replacing p, I, and e by the values given in the problem, 

D = .288 in. 
Consequently, the work of deformation is 

W=-P'D= ^- = 2442 in. Ib. 
2 6 Je 2 

Therefore, from the equality |P'D = P(h + D), we have 

2442 = .75P(6 + .288); 
P = 618 Ib. 



whence 



Problem 138. From what height can a weight of half a ton fall on the middle 
of the beam in the preceding problem without producing a stress greater than 
40,000 lb./in. 2 ? 

75.* Influence line for bending moment. As a load moves over a 
structure the bending moment and shear at any given point change 

continuously. This varia- 
tion of the bending moment, 
shear, or any similar func- 
tion at a given fixed point 
due to a moving load can be 
represented graphically by 
a curve (or straight line) 
called an influence line. 

To obtain the influence 
line for bending moment for 

a simple beam of length /, let d denote the distance of the given point 
A from the left support 0, and x the distance of a movable load P from 




FIG. 77 



(Fig. 77). Then, if P is on the right of A, R^ = 



the moment at A is 






an d hence 



P(l-x)d 
I 



Now let P be a unit load (say one pound or one ton). Then 



* For a brief course the remainder of this chapter may be omitted. 



FLEXUBE OF BEAMS 



97 



load is on the left of A, M a = X ^ ~ ' 



and if the values of M a corresponding to each value of x from d to 
/ are laid off as ordinates, we obtain the straight line A'B', which 
therefore represents the variation in the bending moment at the 
point A as the unit load moves from B to A. Similarly, if the unit 

which is the equation of the 
straight line O'A'. At D 1 both lines have the same ordinate, namely, 

A'E L - The influence line for bending moment is therefore 
l 

the broken line O'A'B'. 

From this construction, it is obvious that the ordinate to the influ- 
ence line at any point D represents the bending moment at A due to 
a unit load at D. Thus, as a 

unit load comes on the beam o - A. _ \B 

from the right, the bending 
moment at A increases from 
the value zero for the load 
at B to the value A'E for the 
load at A, and then decreases 
again to the value zero at 
0. Therefore, having con- 

structed for a unit load the influence line corresponding to any given 
point A, the moment at A due to a load P is found by multiplying 
P by the ordinate to the influence line directly under P. 

Problem 139. Find the position of a system of moving loads on a beam so that 
the bending moment at any point A shall be a maximum. 

Solution. Let (YA'B' be the influence line for bending moment for the point A, 
and let the loads on each side of A be replaced by their resultants PI and P 2 
(Fig. 78). Then, if y l and y z are the ordinates to the influence line directly under 
PI and P 2 , the moment at A is 




FIG. 78 



Now, if the loads move a small distance dx to the left, the moment at A becomes 

M a + dM a = Pi ( yi - dx tan a) + P 2 (y z + dx tan/3). 
Therefore, by subtraction, 

dM a = - PI dx tan a + P 2 dx tan/S, 



and hence 



5 = -Pi tana + P 2 tan/S. 
dx 



98 



STRENGTH OF MATERIALS 



For a maximum value of M a , ^-? = 0, in which case 

dx 



This equation may be written 



P 2 tan/3 = 

p A'C' = 
2 C'B' 

P 2 



A'C' 

O'C'' 



from which, by composition, 



C'B' O'C" 



Pi 

0'C"' 



which is the criterion for maximum moment at A. Expressed in words, the moment 
at any point A is a maximum when the unit load on the whole span is equal to the 
unit load on the smaller segment. 

76. Influence line for shear. To obtain the influence line for 
shear, let /, d, and x have the same meaning as in the preceding 

article. The shear at any 
point A is equal to the re- 
action at 0, and for a unit 
load this reaction is 

I x 



o 




FIG. 79 



I 



If, then, the values of 7^ for 
all values of x from d to I 
are laid off as ordinates, the locus of their ends will be the straight 
line B'A' (Fig. 79). Similarly, for a unit load on the left of A the shear 

/yt 

at A is negative, and its amount is R 2 = > which is the equa- 
tion of the straight line O'A". Since the slopes of the two lines 
A'B' and O'A" are equal, these lines are parallel. The influence line 
for shear is, then, the broken line 0'A"A'I>'. 

As a load comes on the beam from the right the shear at A gradu- 
ally increases from the value zero for the load at B to the value A'E 
for the load just to the right of A. As the load passes A the shear at 
this point suddenly decreases by the amount of the load, thus becom- 
ing negative, and then increases until the load reaches 0, when it 
again becomes zero. Consequently, the shear at A, due to a load P at 



FLEXURE OF BEAMS 



99 




any point C, is found by multiplying P by the ordinate to the influ- 
ence line at C', directly under C. 

Problem 140. Find the position of a system of moving loads on a beam so that 
the shear at any point A shall be a maximum. 

Solution. Let the influence line for the point A be as represented in Fig. 80. 
Also let PI and P 2 be two consecutive loads, d the distance between them, and P' 
the resultant of all the loads on 
the beam. Since A'E is the 
maximum ordinate to the influ- 
ence line, the maximum shear 
at A must occur when one of 
the loads is just to the right of 
A. Suppose the load PI is just 
to the right of A. Then as PI 
passes A the shear at A is sud- 
denly decreased by the amount 
PI. If the loads continue to 
move to the left until P 2 
reaches A, the shear is gradu- 
ally increased by the amount 
P'd tan a, since the ordinate 

under each load is increased by the amount d tan a. Consequently, either P t or 
PZ at A will give the maximum shear at this point according as 



A 



E 




13' 



FIG. 80 



PI ^ P'd tan a ; 



or, since tana = -, according as 



By means of this criterion, it can be determined in any given case which of two 
consecutive loads will give the greater shear at any point. 

77. Maxwell's theorem. When a load is brought on a beam it 
causes every point of the beam to deflect, the amount of this deflec- 
tion for any point being the corresponding ordinate to the elastic 
curve. If, then, a number of loads rest on a beam, the deflection at 
any point of the beam is the sum of the deflections at this point due 
to each of the loads taken separately. 

For example, if two loads P l and P 2 rest on a beam at the points 
A and B respectively, the deflection at one of these points, say A, is 
composed of two parts, namely, the deflection at A due to P l and the 
deflection at A due to P 2 . Similarly, the total deflection at B is com- 
posed of the partial deflections due to P l and P z respectively. 



100 STRENGTH OF MATERIALS 

Maxwell's theorem, when modified so as to apply to beams, states 
that if unit loads rest on a beam at two points I and K, the deflection 
at I due to the unit load at K is equal to the deflection at K due to 
the unit load at L The following simple proof of the theorem is due 
to FoppL* 

Consider a simple beam bearing unit loads at two points / and K 
(Fig. 81). Let the deflection at JTdue to a unit load at /be denoted 
by c7" H , the deflection at / due to a unit load at / by J {i) etc., the 
second subscript in each case denoting the point at which the unit 
load is applied, and the first subscript the point for which the number 
gives the deflection. Thus J ik denotes the influence of a unit load 

at K on the deflection at /. 
For this reason the quantity 
J ik is called an influence num- 
ber. 



If the load at I is of 
amount P it the deflection at 
FIG. 81 / is J a P it that at K is J K P it 

etc. 

Now suppose that a load P { is brought on the beam gradually at 
the point /. Then its average value is ^-P t .,the deflection under the 
load is J a P it and consequently the work of deformation is \ P.(Jf.P^. 
After the load P. attains its full value suppose that a load P k is 
brought on gradually at K. Then the average value of this load is 
\P k , but since P { keeps its full value during this second deflection, 
the work of deformation in this movement is P^J^P^ + J- -^VM^-P*)' 
Therefore the total work of deformation from both deflections is 



Evidently the same amount of work would have been done if the 
load P k had first been applied, and then P t . The expression for the 
total work obtained by applying the loads in this order is 



Therefore, equating the two expressions for the work of deformation, 
which proves the theorem. 

* Festigkeitslehre, p. 197. 



FLEXURE OF BEAMS 



Problem 141. A beam bears a load of 15 tons at a certain point A, and its 
deflections at three other points, B, (7, D, are measured and found to be .30 in., 
. 16 in. , and .09 in. respectively. If loads of 6, 12, and 8 tons are brought on at .B, C, 
and D respectively, find the deflection at A. 

Solution. The deflections at U, C, and D due to a unit load (one ton) at A are 

'- = .02 in., '- - .01 in., and ' = .006 in. respectively. Therefore, by Maxwell's 
15 16 16 

theorem, the deflection at A is 

D a = .02 x 6 + .01 x 12 + .006 x 8 = .268 in. 



78. Influence line for reactions. The most important application 
of Maxwell's theorem is to the determination of the unknown reac- 
tions for a continuous beam. 

Consider a beam continuous over three supports, as shown in 
Fig. 82. Suppose the middle support removed and a unit load (say 1 
ton) placed at this point. A I B a 

Then, if the elastic curve 
is plotted, the ordinate to 
this curve at any point / 
is the deflection at I due 
to the unit load at B, or, 
in other words, this ordi- 
nate is the influence num- 
ber J ib . Similarly, the ordinate to the elastic curve at B is the influence 
number J bb . 

Now -K 2 , the unknown reaction at B, must be of such amount as 
to counteract the deflection at B due to a load P at any point /. 
Therefore 







** 


< IT 


-> 


% ? * ? 

#2 


\ 


S 






But, by Maxwell's theorem, J bi = J^ ; consequently 



The influence numbers J^ and J^ are known as soon as the elastic 
curve for unit load at B is plotted. Therefore, in this case, the con- 
struction of one elastic curve gives sufficient data for all further 
calculations. 



i\fitBBNGTH OF MATERIALS 



Since for any point / the fraction is proportional to J ib (the 

Jbb 
denominator being constant), the elastic curve is called the influence 

line for reactions. 

For a number of concentrated loads P 1? P 2 , -, P n the same method 
applies, R z in this case being given by the equation 



J 



U 



J 



or, more briefly, 



To determine the reactions for a beam continuous over four sup- 
ports and bearing a single concentrated load P at any point /, suppose 
the two middle supports removed. Then if a unit load is placed at B 

(Fig. 83) and the elastic 
curve drawn, the ordinate 
to this curve at any point 
/ is the influence number 
J ib . Similarly, by placing 
a unit load at C and con- 

structing the corresponding 
FIG. 83 

elastic curve, the influence 

number J ic is obtained. Now the reaction R z must be of such amount 
as to counteract the deflections at B due to a load P at / and a load 
R at a Therefore 



A } 


P 

,1 B C D 




* 


., 


#3 



Similarly the reaction JK 3 must be of such amount as to counteract the 
deflections at C due to a load P at I and a load R z at B. Therefore 



By Maxwell's theorem, J bi = J^ and J ci = J ic . Making these substi- 
tutions and solving the above equations simultaneously for J? 2 and jft 3 , 



T> 

2 ~~ 



T-> p 

8= 



FLEXURE OF BEAMS 103 

79. Castigliano's theorem. Consider a beam bearing any number 
of concentrated loads P lf P 2) ---,P n , acting either vertically upward 
or downward, and let W 
denote the work of defor- ' PI \ P * 

mation due to these loads 
(Fig. 84). Then if one of the 
loads, say P., is increased 
by a small amount dP it the 
deflection of P l is increased 
by the amount j; t dP if that FlG> 84 

of P 2 by the amount J 2i dP { , etc., where J u , J 2i , etc., are influence 
numbers. Therefore the work of deformation is increased by the 
amount 

dw = P^dP, + P 2 J 2i dP { + + P n JJP t ; 
whence dw 



In forming this expression the work done by dP { itself has been 
neglected, since it is infinitesimal in comparison with that done by 
P lt P 2) etc. 

Now, from Maxwell's theorem, J ik J ki . Therefore the above expres- 
sion becomes -, w 

= P^ + P 2 J i2 + - - - + P n J in . 

The right member of this equality, however, is the total deflection 
D i at the point /, due to all the loads. Consequently the above expres- 
sion may be written ^ w 

dp, =D> - 

Since the work of deformation W is a function of all the loads and 
not of P. only, this latter expression should be written as a partial 
derivative; thus 



and in this form it is the algebraic statement of Castigliano's theorem. 
Expressed in words, the theorem is : The deflection of the point of 
application of an external force acting on a beam is equal to the par- 
tial derivative of the work of deformation with respect to this force. 



104 



STRENGTH OF MATERIALS 



80. Application of Castigliano's theorem to continuous beams. 

Castigliano's theorem affords still another means of determining the 
unknown reactions of a continuous beam ; for the reactions may be 
included among the loads on the beam, and since the points of applica- 
tion of these reactions are assumed to be fixed, their deflections are zero. 
Therefore, if P k is one of the reactions, D k = 0, and consequently 



o. 



A condition equation of this kind can be found for each reaction, and 
from the system of simultaneous equations so obtained the unknown 
reactions may be calculated. The following problems illustrate the 

application of the theo- 
rem. 

Problem 142. A uniformly 
loaded beam of length 2 1 is 
supported at its center and 
ends. Find the reactions of 
the supports by means of Cas- 
tigliano's theorem. 

Solution. Let w denote the 
unit load on the beam (Fig. 85). 




-P. 



FIG. 85 



From symmetry, PI = P 8 . Also, by taking moments about JB, 

For a point in the first opening at a distance x from the left support, 

~~ 1 2 ' 



consequently, 



W 



=\L 



2 Jo El 2 El I 



uw~\ 

20 J' 



The work of deformation for the other half of the beam is of the same amount. 
Therefore the total work of deformation is 



W ~El[ 8 4 + 20 J' 

Since PI is a function of P 2 , the partial derivative of W with respect to P 2 is 



dTT_j_pPiJ 8 5Pi_w^ 4 aPil 
ap .BiL s '.&PS 4'*aPsj 



FLEXURE OF BEAMS 

Therefore 

By Castigliano's theorem, ^- = 0. Therefore 



105 



dW _ I s Fwl _Pi1 
5P 2 ~ EI\_~S~ ~3~J 



_ 

EI L 8 3 



whence 



Substituting in this expression the value of PI in terms of P 2 , 

P 2 = | wl. 

Problem 143. A uniformly loaded beam extends over three openings of equal 
span. Find the reactions at 
the supports. 

Solution. Let I denote the 
length of each span and w the 
unit load (Fig. 86). 

From symmetry, P x = P 4 
and P 2 = P 3 . Also, by taking 
moments about .B, 

- f 

z FIG. 86 

For any point in the first opening at a distance x.from the left support, 




and therefore, as in the preceding problem, 



W = 



3 



PlWl * 4- ^1 
4 20 J 



Since PI = P 4 , TFhas the same value for the third opening, that is, W% = W\. In 
the second opening 



wx 2 



and therefore 






2 El I 3 3 20 3 4 12 j 

Hence the total work of deformation for all three openings is 



3 



12 



Therefore 
dW 1 



_ 
2-JBIl * P t 4 



17 lot* \ 
12 J" 



106 STRENGTH OF MATERIALS 



Since PI = - P 2 , = 1, and hence 
2 dPz 



e)P 2 

Putting = 0, and substituting for Pj its value in terms of P 2 , 



whence P 2 = 

and consequently PI = f wZ. 

81. Principle of least work. Differentiating partially with respect 

dW 

to P t both members of the equation = D i9 we have 

dP 



As the load increases the deflection increases, and vice versa. There- 

o y~\ 

i 



fore, since dZ> t and dP. have the same sign, i is positive and hence 



. 

- is also positive. But, from the differential calculus, 

dW A ?W^ n 

- = and _>0 

are the conditions that W shall be a minimum. Consequently, the 

reactions of a continuous beam, calculated from the condition = 0, 

ftP< 
are such that they make the work of deformation a minimum. 

In Article 73 it was pointed out that the internal work of defor- 
mation is a form of potential energy. The above is thus a special 
case of what is known as the principle of least work, the general state- 
ment of this principle being expressed by the following theorem : 

For stable equilibrium the potential energy of any system is a 
minimum. 

The importance of the principle of least work is due to the fact 
that it is a general mechanical principle, affording a general solution 
of all problems involving the static equilibrium of elastic solids. Its 
most useful application, perhaps, is to problems which are other- 
wise statically indeterminate, that is to say, problems in which the 



FLEXURE OF BEAMS 107 

number of unknown quantities involved is greater than the number 
of relations furnished by the ordinary conditions of equilibrium. 

The general solution of any problem of this nature by the method 
of least work is as follows : First express the work of deformation 
(or potential energy) in terms of the unknown quantities which it is 
required to determine. Then the condition that this expression shall 
be a minimum resolves itself into the condition that the partial 
derivatives of the potential energy with respect to each of the un- 
knowns involved shall be zero. In this way we obtain exactly as 
many equations as unknowns, from which these unknown quantities 
may be found. 

Thus if W denotes the work of deformation and P v P z , -, P n the 
unknown quantities to be found, first express W 7 " as a function of 
these unknowns, say W(P lt P 2 ,---,P W ). Then the condition for a 
minimum is dW = 0, or, expressing the total differential d W in terms 
of its partial derivatives with respect to the various unknowns, 



.,.. 

Since P Jt P 2 , - , P B are assumed to be independent, in order for this 
relation to be satisfied identically, that is for all values of P lt P z ,>, 
P w , the coefficients of dP v dP 2 , - -,dP n must all be zero ; that is, 



_ 



We have, therefore, n equations from which to determine the n 
unknowns P lf P 2 ,--,P n . 

Before applying this principle it is necessary to find an expression 
for the work of deformation of elastic solids subjected to direct stress 
or to bending stress. 

1. Direct stress. Consider a prismatic bar of length / and cross 
section F, which is subjected to a direct stress, either tension or 
compression, of intensity p. Then from Hooke's law 



108 



STRENGTH OF MATERIALS 



or if P denotes the total load, then since p = and s = - > this 

PI PI 

becomes = E t whence A/ = - If, then, the load is applied 



gradually, the average force acting on the bar during deformation is 
\ P, and consequently the work of deformation in this case is 



2FE 



2. Bending stress. The work of deformation of a prismatic beam 
subjected to a bending moment M has been found in Article 72 to be 



w 



r 



2 El 



C 



The application of the method of least work will now be illustrated 
by a number of simple problems. Problems 142 and 143, Article 80, 
and Articles 82 and 83 are also applications of this principle. 

Problem 144. Three Carnegie I-beams, No. B 80, are placed 4 ft. apart across 
an opening 25 ft. wide. Across their centers is placed another I-beam of the same 

dimensions as the first, and upon 
the center of this cross beam there 
rests a load of 10 tons. Find the 
greatest stress which occurs in 
any member of the construction. 
Solution. Let the amount of 
the load at fl", which is carried by 
GK, be denoted by P (Fig. 87). 

A\ " ' T T Ifl Then the loadg on AB and E p 

J at G and K respectively are each 

p 
-, Q _ equal to , and the load on CD 

-T IG. 81 2 

is 20,000 Ib. - P. 

Now the work of deformation for a simple beam of length I bearing a single 
concentrated load P" at its center is, from Problem 135, 



W = 



Therefore, since the load on AB or EF is , the work of deformation for either 
of these beams is 

P*Zf 
ab ~ * ~~ 384 El ' 



K 


R 


; 


i 


G 


i \ 





, 7_ow' __, 





FLEXUKE OF BEAMS 



109 



Similarly the work of deformation for CD is 






(20,000 - 



and for GK is 



96 El 



96 #1 

Hence the total work of deformation for the entire construction is 
... P*l* , (20,000- 



192 #/ 



96.EI 






By the principle of least work, = 0; consequently 

dP 

dW _ PI* (20,000 - P)l* PI* 



dP 96 El 



48 El 



48 #/ 



=o. 



From the Carnegie handbook, I = 795.6 in.*, and from the figure, li = 300 in., 
Z 2 = 96 in. Inserting these numerical val- 
ues in the above expression, and solving 
for P, p _ 



Having determined P, the stress in the 
various members can easily be calculated. 
Thus it is found that the greatest stress 
occurs in CD, its amount being p = 23,593 
Ib./in.a 

Problem 145. Two short posts of the 
same length I but of cross-section areas FI 
and F 2 and of material having moduli EI 
and E 2 carry a load P jointly. How much 
of the load is carried by each ? (Fig. 88.) 

Solution. Let R denote the load carried 
by No. 1. Then the load carried by No. 2 
is P R. Hence, applying the expression, 
for the work due to a direct stress, the total 
work of deformation for both posts is 

W = 



P-R 



FIG. 88 



The condition for a minimum gives 



whence 



R= 



r. 



FIG. 89- 



Problem 146. A post supporting a load P is braced 
near the bottom by two braces each of length I and inclined 
at the same angle a to the horizontal (Fig. 89). If the 



110 STRENGTH OF MATERIALS 

upright is of cross section FI and has a modulus J? 1 , and the braces are each of 
cross section F 2 and modulus E 2 , show that the load R carried by the upright is 
given by 




Problem 147. A platform 12ft. x 18ft. in size and weighing 1 ton is supported 
at the corners by four wooden legs, each 8 in. square. A load of 5 tons is placed 
on this platform 4 ft. from each of two adjacent edges. How much of the load is 
carried by each leg ? 

Problem 148. A beam 20 ft. long is supported at each end and at a point dis- 
tant 5 ft. from the left end. It carries a load of 180 Ib. at the left end, and of 
125 Ib. at a point distant 6 ft. from the right end. Find the reactions of the 

supports. 

Problem 149. Two beams are supported as shown 
in Fig. 90, the lower beam resting on fixed end sup- 
ports, and the upper beam resting on three supports, 
at its center and ends. The upper beam carries a 
FIG. 90 uniform load. Find the center load transmitted to 

the lower beam. 

Problem 150. A flitched (or composite) beam consists of a 3-in. I-beam weighing 
1\ Ib./ft. and a 4-in. x 6-in. timber, the I-beam being placed underneath the wooden 
beam, and the two are hung from a crane by a wrought-iron strap around the 
middle. A cable is then looped over the ends of this flitched beam 2f ft. distant 
from the center on each side, and a load of 1000 Ib. supported by the loop. Find 
the total load carried by each beam. i 

Problem 151. The king post truss shown in \* --- c 4 
Fig. 91 is formed of a single beam AC resting 
on supports at A and C and trussed at the center 
with a strut BD, supported by two tie rods AD 
and DC. Determine the load R carried by the strut 
BD when a load P is placed at a distance c from A. 

Solution. Let R denote the stress in BD. Then if h denotes the length of the 
strut BD and d the length of each tie, AD and DC, the stress in AD or DC is 

in each, and the direct stress in ABC is -- Let F v F 2 , F 3 , denote the cross- 
2 h 4/i 

section areas of AC, AD, and BD respectively. Then the total work of deforma- 
tion, due to direct stresses in the various members, is 

R 2 h 




In addition to this it is also necessary to consider the work of deformation due to 
the bending stress in AC. At a point distant x from A this is as follows : 

For x between A and P, M AP = [ P(/ ~ c) - -1 x, 
for x between P and B, N PR = PC - (- + \x, 
for x between B and C, M B c = ( \(l x). 



FLEXURE OF BEAMS HI 

Hence the total internal work due to bending is 



Now applying the condition = to the sum of these expressions, and solving 
the resulting equation for #, we have finally 

3d 2 -4c 3 



h d 3 



Problem 152. A wooden beam 12 in. deep, 10 in. wide, and 20 ft. long between 
supports is reenforced by a steel rod 2 in. in diameter and a cast-iron strut 3 in. 
square and 2 ft. high, the whole forming a king post truss. Find the stress in each 
member due to a uniform load of 1200 Ib./ft. over the entire beam. 

82. General formula for flexural deflection. The ordinary method 
of determining flexural deflection is by computing the ordinate to 
the elastic curve at the required point, each case requiring sepa- 
rate treatment. A general formula for flexural deflection, however, 
may be obtained by applying the method of least work in the form 
of Castigliano's theorem. 

From Article 73 the work of deformation due to bending is 



Now in order to apply Castigliano's theorem to this expression, 
assume a concentrated load K applied to the beam at the point 
whose deflection is desired, and let this load be subsequently reduced 
to zero. Let 

M = moment at any section due to given loading, 

M ' = moment at any section due to a unit load at a given point. 

Then for a load K at the given point, the moment at any. section due 
to this load becomes KM ', and hence the total moment due to the 
given loading and the assumed load K is 



112 STRENGTH OF MATERIALS 

v 

Therefore the above expression for the work of deformation now 

1360011168 



By Castigliano's theorem the actual deflection D B due to the given 
loading only is 



and hence applying this to the expression for W B , we have 



2 (M+ KM') 2-(M+KM') 



j 

** 



or, simplifying, 

^MM' 



which is the required general formula for flexural deflection.* All 
the ordinary formulas for the flexural deflection of beams under 
various loadings and with different methods of support are simply 
special cases of this general formula, as illustrated by the following 
examples. 

Problem 153. Find the flexural -deflection at the center of a simple beam of 
constant cross section and bearing a single concentrated load P at the center. 

Px 
Solution. Here Jf x = , and applying a unit load at the point whose deflection 

Cf 

is desired, namely the center, M ' = - . Consequently, 

* 1 

PZ3 



4 El 48 El 

Problem 154. Find the flexural deflection at the center of a simple beam of 
constant cross section bearing a uniform load over the entire span. 

Solution. In this case M = and M ' - Consequently, 

22 2 

i 

dx 



El 384Z 



* This formula is due to Professor Fraenkel, but it is believed that the above proof 
has never before been given. 



FLEXURE OF BEAMS 113 

Problem 155. Find the flexural deflection at the center of a beam of constant 
cross section fixed at both ends and bearing a single concentrated load at the center. 

Solution. The first step in this problem is to determine the moment at one sup- 
port. This is determined from the condition that the deflection at the support 
is zero. 

Applying a load of unity at the left support, we have for sections on either side 
of the center, 

left of center I M = N ~ Y ' right of center I M = M ~ T + P (* ~~ 2~)' 

\^M' = x, LM' = x, 

where M denotes the moment at the left support. Substituting these values in 

the condition 

D B at support = 0, 
we have 



whence 

*.=? 



i pf^ Pan 2 i rn^ Px* P& 2 T 

EI\. 2 " 6 Jo JJL 2 " 6 " 4 Ji 



Proceeding now to find the center deflection, we have 



and, consequently, 

2 r*/PP Pte -Plx'. Px 2 \ , PZ 8 



192^1 

Problem 156. Find the flexural deflection at the center of a beam of constant 

cross section fixed at both ends and bearing a uniform load over the entire span. 

Solution. Let M Q denote the moment at the left support, and w the load in 

>/" Then W ix rf 

M=M --T + -2-' 

M' = z. 
To find Mo apply the condition that the deflection at the support is zero. Then 



at support 



=/<- 



whence 



J_[M< _ W wxn* = 
~ El L 2 "6 8 Jo 



114 STRENGTH OF MATERIALS 

To find the deflection at the center, we have therefore 

wlx wx 2 wl 2 wlx wx 2 



' = --- (from Problem 155), 
8 2 



wl 2 x wlx 2 wl"x wlx 2 wx s \ , wl* 



and, consequently, 

2 r 2 /wl 3 
~ #7 Jo \~96~ 16 ' 10 24 ' 4 4 / ~ 384 El 

83. General formula for shearing deflection. A general formula 
for shearing deflection of beams may also be obtained by the method 
of least work. For this purpose let W s denote the work of deforma- 
tion due to shear, and G the shear modulus. Then if q l denotes the 
unit shearing stress, Hooke's law for shear reads 



and the unit work of shearing deformation for an infinitesimal paral- 
lelepiped of unit volume becomes 



Therefore, since dV=dFdx, the total work of shearing deformation 
for the entire beam is 

TT7 Cj C& dF 

W s = I dx I 

Now to determine the shearing deflection, assume a concentrated 
load K applied to the beam at the point whose deflection is desired, 
and having used K as required by Castigliano's theorem, let it be 
subsequently reduced to zero. 

For this purpose let 

Q = total shear on any variable section due to the given loading, 
Q 1 = shear on any variable section due to a unit load at a given point, 
q = unit shearing stress due to total shear Q as above, 
q 1 = unit shearing stress due to shear Q'. 

Then for a concentrated load K at any given point the shear on any 
section is Q'K, and the unit shear at a variable point due to this load 



FLEXURE OF BEAMS 115 

is q'K. Hence the total unit shear due to both the actual given 
loading and the assumed concentrated load K becomes 



Hence the expression for the work of shearing deformation now becomes 



Now by Castigliano's theorem the actual shearing deflection due to 
the given loading is 



Performing the indicated differentiation and substitution, we have 
therefore 



= dx 



/ 



20 



To simplify this expression, assume the straight-line law of distribu- 
tion of stress, namely = , or q f = q, whence finally 

J Q Q' Q 



which is the required general formula for shearing deflection. The 
method of applying this general formula is illustrated below.* 

Special Case I. Beam of constant rectangular cross section of 
height h. 

From equation (28), Article 56, the unit shear at any point of a 
cross section bounded by parallel sides is 

q= i 

and from equation (29), for a rectangular cross section of li eight li 
this becomes O /k 2 1 2 \ 

I \ 8 2 / 

\ / 

* For applications of this method to beams of variable cross section see article by 
S. E. Slocum, in Journal Franklin Institute, April, 1911. 



116 STRENGTH OF MATERIALS 

Substituting this value of q in the second integral of the general 
formula, it becomes for the special case under consideration 



f)1 />2/I, 

^-/- 



144 # 2 

"8 

Hence the formula for the shearing deflection reduces in this case 
to the simple integral 



Problem 157. Determine the shearing deflection at the center of a simple beam 

of constant rectangular cross section due to a single concentrated load at the center. 

Solution. Let P denote the load at center. Then the total shear on any section is 



Also assuming a unit load at the point whose deflection is to be determined, namely 
the center, we have 

Substituting these values in the above formula, the shearing deflection DS is 

i 

. n . 4^ ZPl 

JL)s = - 



FG WGbh 

To determine the relative amounts of the shearing and flexural deflections, 
assume the relation between the two moduli G and E as G = % E. Then 



3 PI 

4:Ebh 

4Ebh* 



='@ 



Hence the relative value of the shearing and bending deflections depends in this 
case on the square of the ratio of the depth of the beam to its length. Thus 
if h = $l, D s = .12D B ; if h = ^l, D s = .03 AB; if h = -fol, D s = .0075D fi , etc. 
The relative dimensions for which the shearing deflection ceases to be of importance 
are thus easily determined. 



FLEXURE OF BEAMS 117 

Special Case II. Beam of constant circular cross section of radius r. 
From Article 57, the expression for the unit shear, namely 



becomes in the case of a circular cross section 

= Qx* 

Substituting this value of q in the second integral of the general 
formula, we have 



32 Q 2 r r 






9-w-V 9 ^ 
Consequently, the shearing deflection in this case is 



Problem 158. Determine the shearing deflection at the center of a simple beam 
of constant circular cross section and bearing a uniform load. 

Solution. Let the uniform load be of amount w Ib. per unit of length. Then 

Q = wx. Also assuming a unit load at the point whose deflection is to be de- 
termined, namely the center, we have Q = \. Hence in the present case the shear- 
ing deflection at the center is 

i_ 

1 



The relative amount of the shearing deflection as compared with the flexural 
deflection is, in this case, given by the ratio 



PS _3Qirr 2 G. 



384 El 



118 STRENGTH OF MATERIALS 

or assuming, as above, that G = \ E and denoting the depth of the beam by h = 2 r, 
this becomes 

>.s 5 



For a circular cross section, therefore, the shearing deflection is of less relative im- 
portance for a given ratio of depth to length than for a rectangular cross section. 
Thus, if h = ^ I, D s = .016 D , etc. 



EXERCISES ON CHAPTER IV 

Problem 159. In building construction the maximum allowable deflection for 
plastered ceilings is ^ of the span. A floor is supported on 2 in. x 10 in. wooden 
joists, 14 ft. span and spaced 16 in. apart on centers. Find the maximum load 
per square foot of floor surface in order that the deflection may not exceed the 
amount specified. 

Problem 160. Determine the proper spacing center to center for 12-in. steel I- 
beams weighing 35 Ib./ft. for a span of 20 ft. and a uniform floor load of 100 lb./ft. 2 
in order that the deflection shall not exceed ^ of the span. 

Problem 161. One end of a beam is built into a wall and the other end is sup- 
ported at the same level by a post 12 ft. from the wall. The beam carries a uniform 
load of 100 Ib. per linear foot. Find the position and amount of the maximum 
moment and also of the maximum deflection. 

Problem 162. One end of a beam is built into a wall and the other end rests 
on a prop 20 ft. from the wall, at the same level. The beam bears a concentrated 
load of 1 ton at a point 8 ft. from the wall. Find the position and amount of the 
maximum moment and also of the maximum deflection. 

Problem 163. A simple beam of length I carries a distributed load which varies 
uniformly from at one end to w Ib. per unit of length at the other. Find the 
maximum deflection. 

HINT. Note that in the notation of Article 67, 

d*y 

El - = load per unit length, 
dx* 

EI^ = shear, 
dx 3 

d 2 y 

El - moment, 
dx 2 

EI = EIx slope of elastic curve, 
dx 

Ely = El x deflection. 
In the present case, taking the origin at the light end, 



dx* I 
which may be integrated to obtain the deflection. 



FLEXURE OF BEAMS 119 

Problem 164. A beam of uniform strength is one whose moment of resistance 
is in the same constant ratio to the bending moment throughout, so that the skin 
stress is constant. 

Show that in order for a cantilever bearing a single concentrated load at the 
end to be of uniform strength, if the depth is constant, the plan of the beam must 
be triangular ; whereas if the breadth is constant, the side elevation of the beam 
must be parabolic. 

Problem 165. A structural steel shaft 8 in. in diameter and 5 ft. long between 
bearings carries a 25-ton flywheel midway between the bearings. Find the maximum 
deflection of the shaft, considering it as a simple beam. 

Problem 166. A wrought-iron bar 2 in. square is bent to a right angle 4 ft. 
from one end. The other end is then imbedded in a concrete block so that it stands 
upright with the 4 ft. length horizontal. If the upright projects 12 ft. above the con- 
crete and a load of 300 Ib. is hung at the end of the horizontal arm, find the 
deflection at the end of this arm. 

Problem 167. A cantilever of length I is loaded uniformly. At what point of 
its length should a prop be placed, supporting the beam at the same level as the 
fixed end, in order to reduce the bending stress as much as possible, and what 
proportion of the load is then carried by the prop ? 

Problem 168. A continuous beam extends over three spans of 20 ft., 40 ft., and 
30 ft., and carries uniform loads of 3, 1, and 2 tons per linear foot on the three 
spans respectively. Find the danger sections and the reactions of the supports. 

Problem 169. A carriage spring is 2| ft. long and is built up of steel leaves 
each 2 in. wide and f in. thick. How many leaves are required to carry a central 
load of 1000 Ib. with a factor of safety of 4, and what is the deflection under 
this load? 

HINT. Consider the material spread out in the form of a triangle of constant 
depth | in. and varying width, fixed at the base and carrying the load at the apex. 
Also compare with Problem 164. 



CHAPTER V 

COLUMNS AND STRUTS 

84. Nature of compressive stress. When a prismatic piece of 
length equal to several times its breadth is subjected to axial com- 
pression it is called a column, or strut, the word " column " being used to 
designate a compression member placed vertically and bearing a static 
load ; all other compression members being called struts. 

If the axis of a column or strut is not perfectly straight, or if the 
load is not applied exactly at the centers of gravity of its ends, a 
bending moment is produced which tends to make the column deflect 
sideways, or " buckle." The same is true if the material is not per- 
fectly homogeneous, causing certain parts to yield more than others. 
Such lateral deflection increases the bending moment, and conse- 
quently increases the tendency to buckle. A compression member is, 
therefore, in a different condition of equilibrium from one subjected 
to tension, for in the latter any deviation of the axis from a straight 
line tends to be diminished by the stress instead of increased. 

The oldest theory of columns is due to Euler, and his formula is 
still the standard for comparison. Euler's theory, how- 
ever, is based upon the assumptions that the column is 

i perfectly straight, the material perfectly homogeneous, 
and the load exactly centered at the ends, assump- 
tions which are never exactly realized. For practical 
purposes, therefore, it has been found necessary to 
modify Euler's formula- in such a way as to bring it 
into accord with the results of actual experiments, as 
X explained in the following articles. 

85. Euler's theory of long columns. Consider a long 
column subjected to axial loading, and assume that 
the column is perfectly straight and homogeneous, and 
that the load is applied exactly at the centers of gravity of its ends. 

120 



COLUMNS AND STRUTS 121 

Assume also that the ends of the column are free to turn about their 
centers of gravity, as would be the case, for example, in a column 
with round or pivoted ends. 

Now suppose that the column is bent sideways by a lateral force, 
and let P be the axial load which is just sufficient to cause the col- 
umn to retain this lateral deflection when the lateral force is removed. 
Let OX and Y be the axes of X and Y respectively (Fig. 92). Then 
if y denotes the deflection of a point C at a distance x from 0, the 
moment at C is M = Py. Therefore the differential equation of the 
elastic curve assumed by the center line of the column is 



which may be written 

dy? 

dy * 

To integrate this differential equation, multiply by 2 -J- Then 

dx 

2 ^ + 2P ^ = 

dx 2 dx El dx 

and integrating each term, 



where C l is a constant of integration. This equation can now be written 




P ~ y 




Integrating again, 



where <7 2 is also a constant of integration ; whence 




y = 

* This is called an integrating factor and makes each term a perfect differential. See 
Granville's Calculus, pp. 438, 444. 




122 STRENGTH OF MATERIALS 

or, expanding, 

x 



\Eic r / \ P \ 

y = ^-^ sin l x *\Jjj} cos ^ + cos lx xl-^ ) sin C, |. 

Now for convenience let the constants in this integral be denoted 
by A, B, and C respectively ; that is to say, let 

I Tf T/~i I ~t? TC* I T* 

IJ^J.G 1 ~ ^ L&lCj . ^ _ I -r 

Then the general integral becomes 

y = A sin (?# + B cos Cte. 

At the ends and JT, where x = and /, y = 0. Substituting these 
values in the above integral, 

-6 = 0, and ^4 sin Cl = 0. 

Since ^4 and B cannot both be zero, sin Cl = ; whence 

Cl = sin^O = XTT, 

where X is an arbitrary integer. Now let X take the smallest value 
possible, namely 1, and substitute for C its value. Then 



whence 

(48) * 

which is Euler's formula for long columns. 

Under the load P given by this formula the column is in neutral 
equilibrium ; that is to say, the load P is just sufficient to cause it 
to retain any lateral deflection which may be given to it. For this 
reason P is called the critical load. If the load is less than this 
critical value, the column is in stable equilibrium, and any lateral 
deflection will disappear when its cause is removed. If the load 
exceeds this critical value, the column is in unstable equilibrium, and 
the slightest lateral deflection will rapidly increase until rupture 
occurs. 

86. Columns with one or both ends fixed. The above deduction 
of Euler's formula is based on the assumption that the ends of the 



COLUMNS AND STRUTS 



123 



column are free to turn, and therefore formula (48) applies only to 
long columns with round or pivoted ends. 

If the ends of a column are rigidly fixed against turning, the 
elastic curve has two points of inflection, say B and D. From sym- 
metry, the tangent to the elastic curve at the center C 
must be parallel to the original position of the axis of 
the column AE, and therefore the portion AB of the 
elastic curve must be symmetrical with BC, and CD 
with DE. Consequently, the points of inflection, B and 
D, occur at one fourth the length of the column from 
either end. The critical load for a column with fixed 
ends is, therefore, the same as for a column with free 
ends of half the length ; whence, for fixed ends, Euler's 
formula becomes 



(49) 



_P = 




FIG. 93 



Columns with flat ends, fixed against lateral movement, are usually 
regarded as coming under formula (49), the terms " fixed ends " and 
" flat ends " being used interchangeably. 

If one end of the. column is fixed and the other end is free to turn, 
the elastic curve is approximately represented by the line BCDE in 
|p Fig. 93. Therefore the critical load in this case is ap- 

I proximately the same as for a column with both ends 

-Y free, of length BCD, that is, of length equal to f BE 
or |^ I ; whence, for a column with one end fixed and the 
other free, Euler's formula becomes 




(50) 



P = 



approximately. 



87. Independent proof of formulas for fixed ends. 

The results of the preceding article can be established 
independently as follows. 

Suppose both ends of the column fixed against turn- 
ing by a moment M Q at each support. Then the moment 

at any point C, distant x from (Fig. 94), is M = - M + Py, and 

therefore the equation of the elastic curve is 



FIG. 94 



124 



STRENGTH OF MATERIALS 



Proceeding as in Article 85, the general integral of this equation is 
found to be 



y 



A sin ( x 
V 




- 




in which A and 5 are undetermined constants. For x = and /, 
y = 0, and -^ = 0. Therefore, by substituting these values in the 
general integral, the following relations are obtained : 




p / P 

sin I J )-0. 





From these conditions, 

cos ( l \l^} = 1 and sin (KI^) = ; 

whence 



and consequently 





FIG. 95 



which is formula (49) of the preceding article. 

Suppose one end of the column is fixed and the other 
free to turn, and let P h denote the horizontal force neces- 
sary to keep the free end from lateral movement (Fig. 95). 
Then the moment at any point C is M = Py P h x, and 
the equation of the elastic curve is 



The general integral of this equation is 





COLUMNS AND STRUTS 125 

in which A and B are undetermined constants. For x = or l t 
y = ; whence 

B = and A = - 



For x = I, ~ = ; whence 




From the last condition, 



This equation is of the form u = tan u, and from this it is found by 
trial that 



= 4.49. 

Consequently, 

20 El 2 -rr^EI 
P = = , approximately. 

V v 

This equation is of the same form as formula (50) of the preceding 
article, the difference between the numerical constants in the two 
formulas being due to the approximate nature of the solution given 
in Article 86. 

88. Modification of Euler's formula. It has been found by ex- 
periment that Euler's formula applies correctly only to very long 
columns, and that for short columns or those of medium length it 
gives a value of P considerably too large. 

Very short columns or blocks fail solely by crushing, the tendency 
to buckle in such cases being practically zero. Therefore, if p denotes 
the crushing strength of the material and F the area of a cross 
section, the breaking load for a very short column is P = pF. * 

* As Euler's formula is based upon the assumption that the column is of sufficient 
length to buckle sideways, it is evident a priori that it cannot be applied to very short 
columns in which this tendency is practically zero. Thus, in formula (48), as I ap- 
proaches zero P approaches infinity, which of course is inadmissible. 



126 STRENGTH OF MATERIALS 

For columns of ordinary length, therefore, the load P must lie 
somewhere between pF and the value given by Euler's formula. 
Consequently, to obtain a general formula which shall apply to 
columns of any length, it is only necessary to express a continuous 

2 jfij 

relation between pF and - Such a relation is furnished by the 
equation 

(51) !>= 

1+pF 



For when / = 0, P = pF, and when / becomes very large P approaches 

2 37*7" 

the. value Moreover, for intermediate values of I this formula 

L 

gives values of P considerably less than given by Euler's formula, thus 
agreeing more closely with experiment. 

89. Rankine's formula. Although the above modification of 
Euler's formula is an improvement on the latter, it does not yet 
agree closely enough with experiment to be entirely satisfactory. 
The reason for the discrepancy between the results given by this 
formula and those obtained from actual tests is that the assumptions 
upon which the formula is based, namely, that the column is perfectly 
straight, the material perfectly homogeneous, and the load applied 
exactly at the centers of gravity of the ends, are never actually 
realized in practice. 

To obtain a more accurate formula, two empirical constants will 
be introduced into equation (51). Thus, for fixed ends, let 



(52) P = 



y\ 2 

v . 



where /and g are arbitrary constants to be determined by experiment, 
and t is the least radius of gyration of a cross section of the column. 
This formula has been obtained in different ways by Gordon, Ran- 
kine, Navier, and Schwarz.* Among German writers it is known as 

* Rankine's formula can be derived independently of Euler's formula either by 
assuming that the elastic curve assumed by the center line of the column is a sinusoid, 
or by assuming that the maximum lateral deflection D at the center of the column is 

72 

given by the expression D = n , where I is the length of the column, b its least width, 
and n an empirical constant. & 



COLUMNS AND STEUTS 127 

Schwarz' formula, whereas in English and American text-books it is 
called Rankine's formula. 

For / = 0, P = gF, and since short blocks fail by crushing, g is 
therefore the ultimate compressive strength of the material. 

For different methods of end support Eankine's formula takes 
the following forms. 

Flat ends, | = - _. 

(fixed in direction) 1 _|_ f I - 



Round ends, 






(direction not fixed) 1 _i_ A fl_ 

J 



Hinged ends, 

(position fixe 
direction) 

One end flat and the other round, 



/ 

, . - 

direction) 



(position fixed, but no f . -^ 1 i O -f I 

L ~T &J ( ~ 



90. Values of the empirical constants in Rankine's formula. 

The values of the empirical constants, / and g, in Eankine's formula 
have been experimentally determined by Hodgkinson and Christie 
with the following results. 

For hard steel, g = 69,000 lb./in. 2 , / = 

For mild steel, g = 48,000 lb./in. 2 , / = 



30,000 
For wrought iron, g = 36,000 lb./in. 2 , / = ^^ 

For cast iron, g = 80,000 lb./in. 2 , / = 

For timber, g= 7,200 lb./in. 2 , / = 

These constants were determined by experiments upon columns for 
which 20<-<200, and therefore can only be relied upon to 



128 STRENGTH OF MATERIALS 

furnish reliable results when the dimensions of the column lie within 
these limits. 

As a factor of safety to be used in applying the formula, Rankine 
recommended 10 for timber, 4 for iron under dead load, and 5 for 
iron under moving load. 

Problem 170. A solid, round, cast-iron column with flat ends is 15 ft. long and 
6 in. in diameter. What load may be expected to cause rupture ? 

Problem 171. A square wooden post 12 ft. long is required to support a load 
of 15 tons. With a factor of safety of 10, what must be the size of the post ? 

Problem 172. Two medium steel Cambria I-beams, No. B 25, weighing 25.25 
lb./ft., are joined by lattice work to form a column 25 ft. long. How far apart 
must the beams be placed, center to center, in order that the column shall be of 
equal strength to resist buckling in either axial plane ? 

Problem 173. Four medium steel Cambria angles, No. A 101, 3 in. by 5 in. in 
size, have their 3-in. legs riveted to a f-in. plate so as to form an I-shaped built 
column. How wide must the plate be in order that the column shall be of equal 
strength to resist buckling in either axial plane ? 

91. Johnson's parabolic formula. From the manner in which 
equation (51) was obtained and afterward modified by the intro- 
duction of the empirical constants / and g, it is clear that Rankine's 
formula satisfies the requirements for very long or very short col- 
umns, while for those of intermediate length it gives the average 
values of experimental results. A simple formula which fulfills 
these same requirements has been given by Professor J. B. Johnson, 
and is called Johnson's parabolic formula. 

If equation (52) is written 



and then y is written for p, and x f or - > Rankine's formula becomes 

t 

y 

For this cubic equation Johnson substituted the parabola 

y = S - ear 2 , 

in which x and y have the same meaning as above, and S and e are 
empirical constants. The constants 8 and e are then so chosen that 



COLUMNS AND STRUTS 



129 



the vertex of this parabola is at the elastic limit of the material 
on the axis of loads (or F-axis), and the parabola is also tangent 
to Euler's curve. In this way the formula is made to satisfy the 
theoretical requirements for very long or very short columns, and 
for those of intermediate length it is found to agree closely with 
experiment. 

For different materials and methods of end support Johnson's 
parabolic formulas, obtained as above, are as follows : 



KIND OF COLUMN 


FORMULA 


LIMIT FOB USE 


Mild steel 






Hinged ends 


- =42,000-. 97 /-V 


-^150 
* ^ 


Flat ends 


- = 42,000 - .62 /-Y 


1^190 
i 


Wrought iron 






Hinged ends 


- = 34,000 - .67 (-\* 


|^170 


Flat ends 


- = 34,000 - .43 (-Y 


i-210 




F \t/ 


t 


Cast iron 








P 25 //\ 2 


I 


Hound ends 


- = 60,000- -(-) 






T> Q / 7\ 2 


I 


Flat ends 


- = 60,000- -(-) 
F 4\tJ 


-^120 

t ^ 


Timber (flat ends) 






White pine 


-= 2,500- .$(-* 


j* 6 


Short-leaf yellow pine 


-= 3,300- -7/-V 


-^60 


Long-leaf yellow pine 


P /Z\ 2 
f =4,000 -.8(1) 


-^ 60 


White oak 


P //\ 2 

-= 3,500- ,8^-J 


-^ 60 



The limit for use m each case is the value of x I = - J at the point 
where Johnson's parabola becomes tangent to Euler's curve. For 

greater values of - Euler's formula should therefore be used. 
t 

* In the formulas for timber if is the least lateral dimension of the column. 



130 



STRENGTH OF MATERIALS 



A graphical representation of the relation between Euler's formula, 
Rankine's formula, J. B. Johnson's parabolic formula, and T. H. John- 
son's straight-line formula (considered in the next article) is given 
in Fig. 96, for the case of a wrought-iron column with hinged ends.* 




300 



FIG. 96. Wrought-iron Column (pin ends) 

1, Euler's formula; 2, T. H. Johnson's straight-line formula; 3, J. B. Johnson's 
parabolic formula ; 4, Rankine's formula 

Problem 174. A hollow wrought-iron column with flat ends is 20 ft. long, 7 in. 
internal diameter, and 10 in. external diameter. Calculate its ultimate strength by 
Rankine's and Johnson's formulas, and compare the results. 

Problem 175. Compute the ultimate strength of the built column in Prob- 
lem 172 by Rankine's and Johnson's formulas, and compare the results. 

92. Johnson's straight-line formula. By means of an exhaustive 
study of experimental data on columns, Mr. Thomas H. Johnson has 
shown that for columns of moderate length a straight line can be 
made to fit the plotted results of column tests as exactly as a curve. 
He has therefore proposed the formula 

* For a more extensive comparison of these formulas see Johnson's Framed Structures, 
8th ed., 1905, pp. 159-171; also Trans. Artier. Soc. Civ. Eng., Vol. XV, pp. 518-536. 



COLUMNS AND STRUTS 



131 



(53) 



P I 

= V (7 

F t 



or, in the notation of the preceding article, 



in which v and <r are empirical constants, this being the equation of 
a straight line tangent to Euler's curve. This formula has the merit 
of great simplicity, the only objection to it being that for short 
columns it gives a value of P in excess of the actual breaking load. 
The relation of this formula to those which precede is shown in 
Fig. 96. 

The constants v and cr in formula (53) are connected by the relation 



where for fixed ends n = 1, for free ends n 4, and for one end fixed 
and the other free n = 1.78. 

The table on page 132 gives the special forms assumed by John- 
son's straight-line formula for various materials and methods of end 
support.* 

The limit for use in this case is the value of x I = - ) for the point 
at which Johnson's straight line becomes 
tangent to Euler's curve. 



^ 



Problem 176. Compute the ultimate strength of 
the column in Problem 104 by Rankine's and John- 
son's straight-line formulas, and compare the results. 

Problem 177. A column 18 ft. long is formed by 
joining the legs of two Carnegie steel channels, No. C 3, 
weighing 30 lb./ft, by two plates each 10 in. wide 
and in. thick, as shown in Fig. 97. Find the safe 
load for this column by Johnson's straight-line for- 
mula, using a factor of safety of 4. 

Problem 178. A wrought-iron pipe 10 ft. long, 
and of internal and external diameter 3 in. and 
4 in. respectively, bears a load of 7 tons. What is the factor of safety ? 



FIG. 97 



* Trans. Amer. Soc. Civ. Eng., 1886, p. 530. 



132 



STRENGTH OF MATERIALS 



KIND OF COLUMN 


FORMULA 


LIMIT FOR USE 


Hard steel 








P I 


7 


Flat ends 


- = 80,000 - 337 - 


- ^ 158.0 




F t 


t 




P I 


j 


Hinged ends 


= 80,000 - 414 - 
F t 


- ^ 129.0 
t 




P I 


j 


Round ends 


- = 80,000 - 534 - 


-^ 99.9 


Mild steel 






Flat ends 


- = 52,500 - 179 - 


-p 195.1 




F t 


t 




P I 


I 


Hinged ends 


- = 52,500 - 220 - 
F t 


- ^ 159.3 


Round ends 


- = 52,500 - 284 - 


- ^ 123.3 




F t 


t 


Wrought iron 


P I 


i 


Flat ends 


- = 42,000 - 128 - 


- ^ 218.1 




F t 


t 




P I 


I _ 


Hinged ends 


- = 42,000 - 157 - 
F t 


t < 




P I 


i 


Round ends 


- = 42,000 - 203 - 


- ^ 138.0 




F t 


t <: 


Cast iron 








P I 


7 


Flat ends 


- = 80,000 - 438 - 


- ^ 121.6 




F t 


t 




P 1 


j 


Hinged ends 


- = 80,000 - 537 - 


-^ 99.3 




F t 


t 




P I 


I 


Round ends 


- = 80,000 - 693 - 


-P 77.0 




F t 


I ^ 


Oak 








P J 


i 


Flat ends 


- = 5,400 - 28 - 


-^128.1 




F t 


t 



93. Cooper's modification of Johnson's straight-line formula. In 

his standard bridge specifications, Theodore Cooper has adopted John- 
son's straight-line formulas, modifying them by the introduction of 
a factor of safety. Thus, for medium steel, Cooper specifies that the 
following formulas shall be used in calculating the safe load. 

For chords 

^ = 8,000 - 30 - for live load stresses, 

JJ t 

P I 

= 16,000 60 - for dead load stresses. 
F t 



COLUMNS AND STRUTS 133 

For posts 

^ = 7,000 - 40 - for live load stresses, 
F t 

j, = 14,000 - 80 - for dead load stresses, 

P I 

- 10,000 60 - for wind stresses. 

_r t 

For lateral struts 

~ = 9,000 - 50 - for initial stresses. 

By initial stress in the last formula is meant the stress due to the 
adjustment of the bridge members during construction. 

Problem 179. What must be the size of a square steel strut 8 ft. long, to trans- 
mit a load of 6 tons with safety ? 

Problem 180. Design a column 16 ft. long to be formed of two channels joined 
by two plates and to support a load of 20 tons with safety. 

Problem 181. Using Cooper's formula for live load, design the inclined end 
post of a bridge which is 25 ft. long and bears a load of 30 tons, 'the end post to be 
composed of four angles, a top plate, and two side plates. 

94. Beams of considerable depth. When narrow beams of consider- 
able depth are subjected to compression, as, for example, in a deck 
plate girder bridge, the strain is similar to that in a column. For a 
narrow, deep beam the inertia ellipse is greatly elongated, and conse- 
quently the radius of gyration relative to a line forming a small angle 
with the horizontal is considerably less than the semi-major axis of 
the ellipse. Therefore, if the beam is thrown slightly out of the ver- 
tical by the unequal settling of its supports, or by any other cause, 
such inclination results in a notable decrease in its resistance. Since 
it is impossible to make allowances for such accidental reductions of 
strength, beams of great depth or very thin web should be avoided. 

95. Eccentrically loaded columns. In a column carrying an eccen- 
tric load, as, for example, a column carrying a load on a bracket, or 
the post of a crane, there is a definite amount of bending stress due 
to the eccentricity of the load in addition to the column stress. As 
the nature of column stress is such that it is impossible to deter- 
mine its amount, the simplest method of handling a problem of this 



134 



STRENGTH OF MATERIALS 



H\ 

pj+P 



kind is to determine its relative security against failure as a column 
and failure by bending. That is to say, first determine its factor of safety 
against failure as a column under the given column 
load. Then consider it as a beam and find the equiv- 
alent bending moment which would give the same 
factor of safety. Finally, combine this equivalent 
bending moment with that due to the eccentric 
load, and calculate the unit stress from the ordinary 
beam formulas. 

To illustrate the method, suppose that a column 
18 ft. long is composed of two 12-in. I-beams, each 
weighing 40 lb./ft., and carries a column load of 20 
tons at its upper end and also an eccentric load of 
10 tons with eccentricity 2 ft., as shown in Fig. 98. 
Assuming that the column has flat ends, and using Johnson's 

straight-line formula P=jP( 52,500 179 -\ the factor of safety 
against column failure is 

^52,500- 179 -\ 
Factor of safety, 6Q?000 

_ 2(11.76) (52,500 - 179(47.3)) _ 
60,000 

Now consider the column as a beam, and find the equivalent central 
load K corresponding to the factor of safety just found, namely 17.3. 
The maximum moment in a simple beam bearing a concentrated load 



FIG. 



K at the center is M= 

4 



Hence from the beam formula M 



we 



, Kl pi . 
have -=-, whence 
4 e 



le 



Assuming the ultimate strength of the material to be 60,000 lb./in. 2 , 
we have 

60,000 



P ~~ 



17.3 
= 216 in., 



lb./in. 2 , 1= 2 (245.9) in. 4 , 
e = 6 in., 



COLUMNS AND STKUTS 135 

and inserting these values, the equivalent load K is found to be 

4 x 60,000 x 491.8 
17.3 x 216 x 6 

Now the eccentric load P 2 acting parallel to the axis of the column 
produces the same bending effect as a horizontal reaction H at either 
end, where HI = P 2 d. The bending moment at the center, due to a 

TT7 

reaction H perpendicular to the axis of the beam, is, however, 

Hence the total equivalent moment at the center now becomes 

M- + - ^- 

_5220x 216 20,000 x 24 

4 2 

= 521,880 in. Ib. 

Consequently, the maximum unit stress in the member becomes 

M 



521,880 , Q ,_ ., .. 2 
== ^l96- ==63671b - /in - 

which corresponds to a factor of safety of about 9. 

If this factor of safety is larger than desired, assume a smaller 
I-beam and repeat the calculations. 

A method substantially equivalent to the above is to assume that 
the stress in a column is represented by the empirical factor in the 
column formula used. Thus for a short block, the actual compressive 
stress p is given by the relation P = pF, whereas in the column 

formula used above, namely P = W52,500 179 - ), the stress p is 

replaced by the empirical factor 52,500 179-. Consequently, the 
fraction 

52,500-179- 
t 



136 



STRENGTH OF MATERIALS 



where u c denotes the ultimate compressive strength of the material, 
represents the reduction in strength of the member due to its slim- 
ness and method of loading ; or, what amounts to the same thing, the 
equivalent unit stress in the column is 



52,500- 179- 
t t 

Applying this method to the numerical problem given above, we 
have ^=23.52, 



- = 47.3, and 



60,000 



52,500 -179 



179 x 47.3 



Hence the equivalent stress in the column is 
30 x 2000 



Pe 



23.52 



x 1.36 = 3470 lb./in. s 



Also, the bending stress, produced by the eccentricity of the load, is 



Consequently, by this method, the total stress in the column is found 

3470 + 2928 = 6398 lb./in. 2 
If a formula of the Rankine-Gordon type is used, namely, 

P_ 9 



the equivalent stress p e in the column, due to the given load P, is 



u. 



9 

where u c denotes the ultimate compressive strength of the material, 
as above. 



COLUMNS AND STKUTS 



137 



EXERCISES ON CHAPTER V 

Problem 182. A strut 16 ft. long, fixed rigidly at both ends, is needed to sup- 
port a load of 80,000 Ib. It is to be composed of two pairs of angles united with 
a single line of -in. lattice bars along the central plane. Determine the size of 
the angles for a factor of safety of 5. 

Note that the angles must be spread in. to admit the latticing. 

Problem 183. For short posts or struts, such as are ordinarily used in building 
construction, it is customary to figure the safe load as 12,000 lb./in. 2 of cross-section 
area for lengths up to 90 times the radius of 

gyration,* i.e. for -^90. To what factor of sp H= 14 1 



S'Toiis 



FIG. 99 



safety does this correspond, using Johnson's 
straight-line formula ? 

Problem 184. The posts used to support a 
girder in a building are 8 in. x 8 in. timbers 
8 ft. long. Find the diameter of a solid cast- 
iron column of equal strength. 

If a wrought-iron pipe 4 in. in external 
diameter is used, what must be its thickness 
to be equally safe ? _ 

Problem 185. At what ratio of diameter 

to length would a round mild steel strut have the same tendency to crush as 
to buckle? 

Problem 186. A load of 100 tons is carried jointly by three cast-iron columns 
20 ft. long. What saving in material will be effected by using a single column 
instead of three, the factor of safety to be 15 in both cases ? 

Problem 187. Determine the proper size for a hard-steel piston rod 48 in. long 
for a piston 18 in. in diameter and a steam pressure of 80 lb./in. 2 Consult table 
for proper factor of safety. 

Problem 188. The side rod of a locomotive is 9 ft. long between centers, 4 in. 
deep, and 2 in. wide. The estimated thrust in the rod is 12 tons, and the transverse 
inertia and gravity load 20 Ib. per inch of length. Determine the factor of safety. 

Problem 189. The vertical post of a crane, sketched in Fig. 99, is to be made 
of a single I-beam. The post is pivoted at both ends so as to revolve about its axis. 
Find the size of I-beam required for factor of safety of 4, and for dimensions 
and loading as shown in the figure. 



CHAPTER VI 

TORSION 

96. Circular shafts. When a uniform circular shaft, such as 
shown in Fig. 100, is twisted by the application of moments of" oppo- 
site signs to its ends, every straight line AB parallel to its axis is 
deformed into part of a helix, or screw thread, A C. The strain in this 
case is one of pure shear and is called torsion, as mentioned in Arti- 
cle 37. The angle <f> is called the angle of shear (compare Article 33), and 
is proportional to the radius BD of the shaft. The angle 6 is called the 
angle of twist, and is proportional to the length AB of the shaft. 





FIG. 100 



FIG. 101 



97. Maximum stress in circular shafts. Consider a section of 
length dx cut from a circular shaft by planes perpendicular to its 
axis (Fig. 101). Let dO denote the angle of twist for this section. 
Then, since the angle of twist is proportional to the length of the 
shaft, dO : 6 = dx : I ; whence 

de-e 

I ' 
Also, if $ and dO are expressed in circular measure, 



and 
Therefore 



. rdO 6 

m = = T 

dx I 
138 



TOKSION 139 

From Hooke's law (Article 33), = G. Hence 

9 

(54) q = G<t> = 

i 

Therefore q is proportional to r ; that is to say, the unit shear is pro- 
portional to its distance from the center, being zero at the center 
and attaining its maximum value at the circumference. 

If q' denotes the intensity of the shear at the circumference and 
a denotes the radius of the shaft, then the shear q at a distance r 
from the center is given by the formula 

q'r 

Let M t denote the external twisting moment. Then, since M t must 
be equal to the internal moment of resistance, 

M t 



t = CqrdF= Cr*dF= 
J aj a 



where I p is the polar moment of inertia of the section. 

For a solid circular shaft of diameter D, I p - r - , and consequently 



For a hollow circular shaft of external diameter D and internal 
diameter d y I p = ~ (D 4 d 4 ), and hence 

(56) q 1 = 



98. Angle of twist in circular shafts. From equation (54), 

Gr Ga' 

Therefore, for a solid circular shaft, from equation (55), 

(57) 6. 



140 STRENGTH OF MATERIALS 

and for a hollow circular shaft, from equation (56), 



If M t is known and can be measured, equations (57) and (58) can 
be used for determining G. If G is known and 6 measured, these 
equations can be used for finding M t ; in this way the horse power 
which a rotating circular shaft is transmitting can be determined. 

Problem 190. A steel wire 20 in. long and .182 in. in diameter is twisted by 
a moment of 20 in. Ib. The angle of twist is then measured and found to be 
9 = 18 31'. What is the value of G determined from this experiment ? 

Problem 191. If the angle of twist for the wire in Problem 190 is 6 40, how 
great is the torsional moment acting on the wire ? 

99. Power transmitted by circular shafts. Let H be the number 
of horse power transmitted by the shaft, and n the number of revolu- 
tions it makes per minute. Then, if q is the force acting on a particle 
at a distance r from the center, the moment of this force is qr, and 

consequently the total moment transmitted by the shaft is M t = I qrdF. 

Also, the distance traveled by q in one minute is 2 Trrn, and there- 
fore the total work transmitted by the shaft is 



W= Cz 



Since 1 horse power = 33,000 ft. Ib./min. = 396,000 in. Ib./min., the 
total work done by the shaft is 

W= 396,000 H in. Ib./min. 
Therefore 

2 irn CrqdF = W= 396,000 H, 

2 im,M t = 396,000 H-, 



. 

2 Trn 

Therefore, if it is required to find the diameter D of a solid circular 
shaft which shall transmit a given horse power H with safety, then 
from equation (55), ^ ^ 16Jff _ 321,000 IT 

~ 



TORSION 



141 



whence 

(59) 



Z> = 68.5 



.-^ 



As safe values for the maximum unit shear q 1 Ewing recommends 
9000 lb./in. 2 for wrought iron, 13,500 lb./in. 2 for steel, and 4500 lb./in. 2 
for cast iron.* Inserting these values of q' in formula (59), it becomes 



(60) 



where for steel p 2.88, for wrought iron p = 3.29, and for cast 
iron IJL = 4.15. 

Expressed in kilowatts in- 
stead of horse power, this 
formula becomes 




where for steel p = 3.175, for 
wrought iron p = 3.627, and 
for cast iron p = 4.576. 

Problem 192. Find the diameter of a solid wrought-iron circular shaft which is 
required to transmit 150 H.P. at a speed of 60 revolutions per minute. 

Problem 193. A steel shaft is required to transmit 300 H.P. at a speed of 200 
revolutions per minute, the maximum moment being 40 per cent greater than the 
average. Find the diameter of the shaft. 

Problem 194. Under the same conditions as in Problem 193, find the inside 
diameter of a hollow circular shaft whose outside diameter is 6 in. Also com- 
pare the amount of metal in the solid and hollow shafts. 

Problem 195. How many H.P. can a hollow circular steel shaft of 15 in. exter- 
nal diameter and 11 in. internal diameter transmit at a speed of 50 revolutions per 
minute, if the maximum allowable unit stress is not to exceed 12,000 lb./in. 2 ? 

100. Combined bending and torsion. When a shaft transmits power 
by means of a crank or pulley, it is subjected to combined bending 
and torsion. For example, if a force P acts at a point A in the crank 
pin shown in Fig. 102, the bending moment at any point C of the 
shaft is M b = Pd^ and the torsional moment at C is M t = Pd r 



Ewing, The Strength of Materials, p. 190. 



142 STRENGTH OF MATERIALS 

Therefore, if D is the diameter of the shaft at (7, the normal stress 
on the extreme fiber due to bending is 

_'32M h 
P ~~~'^^ > 

and the shearing stress on the extreme fiber due to torsion is 

= 16Jf, 
q " TrD 3 ' 

There is also a shearing stress of amount P distributed over the 
cross section through (7, but since it is zero at the outer fiber, it does 
not enter into this calculation. 

From Article 26, the values of the principal stresses are 



and from Article 28, the maximum or minimum shear is 



Inserting in these expressions the values of p and q obtained above, 
the principal stresses and the maximum or minimum shear are, in 
the present case, 

V 'Ml + M\) (called Rankine's formula), 

mill 



16 



</'iiiax = =*= -^Ml + Mf (called Guest's formula). 

mill IT I) 

The equivalent stress may also be found. Thus, from equation (15), 
Article 36, assuming m = 3J, its value is found to be 



16 i 

p e = [.7 M b 1.3 VJWrJ + Ml~] (called St. Venant's formula). 

It is evident that St. Venant's formula is simply a refinement on 
Rankine's, as both give the principal normal stresses, whereas Guest's 
formula is essentially different, since it gives shear. Hence in design- 
ing members subjected to both bending and torsion, try both Guest's 
and Rankine's (or St. Venant's) formulas with the same factor of 
safety, and then use whichever gives the larger dimensions to the 
construction. 



TORSION 143 

Problem 196. A steel shaft 5 in. in diameter is driven by a crank of 12-in. 
throw, the maximum thrust on the crank being 10 tons. If the outer edge of the 
shaft bearing is 11 in. from the center of the crank pin, what is the equivalent 
stress in the shaft at this point ? 

Problem 197. A steel shaft 10 ft. long between bearings and 4 in. in diameter 
carries a pulley 14 in. in diameter at its center. If the tension in the belt on this 
pulley is 250 lb., and the shaft makes 80 revolutions per minute, what is the maxi- 
mum stress in the shaft and how many H.P. is it transmitting ? 

*101. Resilience of circular shafts. In Article 74 the resilience of 
a body was denned as the internal work of deformation. For a solid 
circular shaft this internal work is 



where N t is the external twisting moment and 6 is the angle of twist. 

From equation (54), 6 = -^ = , and from equation (55), M = -- ~- 
Gr Ga 16 

Therefore the total resilience of the shaft is 



2 
and consequently the mean resilience per unit of volume is 

w - W ~ * 
WI ~V"-TG' 

102. Non-circular shafts. The ahove investigation of the distribu- 
tion and intensity of torsional stress applies only to shafts of circular 
section. For other forms of cross section the results are entirely dif- 
ferent, each form having its own peculiar distribution of stress. 

For any form of cross section whatever, the stress at the boundary 
must be tangential. For if the stress is not tangential, it can be 
resolved into two components, one tangential and the other normal 
to the boundary ; and in Article 23 it was shown that such a normal 
component would necessitate forces parallel to the axis of the shaft, 
which are excluded by hypothesis. 

Since the stress at the boundary must be tangential, the circular 
section is the only one for which the stress is perpendicular to a 
radius vector. Therefore the circular section is the only one to which 
the above development applies, and consequently is the only form of 

* For a brief course the remainder of this chapter may be omitted. 



144 STEENGTH OF MATERIALS 

cross section for which Bernoulli's assumption holds true. That is to 
say, the circular section is the only form of cross section which remains 
plane under a torsional strain. 

The subject of the distribution of stress in non-circular shafts has 
been investigated by St. Yenant, and the results of his investigations 
are summarized below (Articles 103-106 inclusive). 

103. Elliptical shaft. For a shaft the cross section of which is an 
ellipse of semi-axes a and b, the maximum stress occurs at the ends 
of the minor axis, instead of at the ends of the major axis, as might 
be expected. The unit stress at the ends of the minor axis is given 

by the formula ,.. 

22 JxL t 



and the angle of twist per unit of length is 



The total angle of twist for an elliptical shaft of length I is therefore 



Problem 198. The semi-axes of the cross section of an elliptical shaft are 3 in. 
and 5 in. respectively. What is the diameter of a circular shaft of equal strength? 

104. Rectangular and square shafts. For a shaft of rectangular 
cross section the maximum stress occurs at the centers of the longer 
sides, its value at these points being 

(61) g max = _ = .68 + .45 



in which h is the longer and I the shorter side of the rectangle. The 
angle of twist per unit of length is, in this case, 



For a square shaft of side b these formulas become 



(62) 

and 



TORSION 145 

The value of q for a square shaft found from this equation is about 

Mr 

15 per cent greater than if the formula q = was used, and the 

p 
torsional rigidity is about .88 of the torsional rigidity of a circular 

shaft of equal sectional area. 

Problem 199. An oak beam 6 in. square projects 4 ft. from a wall and is acted 
upon at the free end by a twisting moment of 25,000 ft. Ib. How great is the angle 
of twist ? 

105. Triangular shafts. For a shaft whose cross section is an 

equilateral triangle of side c, 

M. 

= 20 



and the angle of twist per unit of length is 



The torsional rigidity of a triangular shaft is therefore .73 of the tor- 
sional rigidity of a circular shaft of equal sectional area. 

106. Angle of twist for shafts in general. The formula for the 
angle of twist per unit of length for circular and elliptical shafts can 
be written 

= 

1 G 

in which I p is the polar moment of inertia of a cross section about 
its center, and F is the area of the cross section. This formula is 
rigorously true for circular and elliptical shafts, and 
St. Venant has shown that it is approximately true 
whatever the form of cross section. 

Problem 200. Compare the angle of twist given by St. Venant's 
general formula with the values given by the special formulas in 
Articles 103, 104, and 105. 

Problem 201. Find the angle of twist in Problem 193. 

Problem 202. Find the angle of twist in Problem 194, and com- 
pare it with the angle of twist for the solid shaft in Problem 201. 

107. Helical springs. The simplest form of a helical, 
or spiral, spring is formed by wrapping a wire upon a 
circular cylinder, the form of such a spring being that 

of a screw thread. Let r be the radius of the coil and a the radius 
of the wire, and let the spring be either compressed or extended by 




146 STRENGTH OF MATERIALS 

two forces P acting in the direction of the axis of the cy Under (Fig. 103). 
Then the bending moment at any point of the spring is M = Pr, If 
the radius r of the coil is large in comparison with the diameter of 
the wire, and if the spring is closely wound, the plane of the external 
moment M is very nearly perpendicular to the axis of the helix, and 
consequently the bending strain can be assumed to be zero in com- 
parison with the torsional strain. Under this assumption the maxi- 
mum stress is found, from equation (55), to be 



Tra 7TCT 

Similarly, the maximum stress in a spring of square or rectangular 
cross section can be found by substituting M= Pr in equations (61) 

and (62). 

To find the amount by which the spring 
is extended or compressed, let d6 be the 
angle of twist for an eleruent of the helix 
of length dl. Then (Fig. 104), if AB is the 
axis of the spring, a point M in this axis 
in the same horizontal plane with the ele- 
ment dl is displaced vertically an amount 

MN rdO in the direction of the axis. Therefore the total axial 
compression or extension D of the spring is the sum of all the infini- 
tesimal displacements rdO for every element dl ; whence 

D= CrdO. 




v ,. /Kn a 2 Ml 2Prl 

From equation (57), 6 = = - 

9 Pr 
Therefore dO = dl } and consequently 

TTCl G 

2Pr 2 l 



/ l 2 Pr 2 2 Pr 2 C l 2 Pr 2 < 
ITCH (JT TTCt' (JT Jr. TTCL (j~ 

^ <J 



D 

in which / is the length of the helix. 

If n denotes the number of turns of the helix, then, under the 
above assumption that the slope of the helix is small, I = 27rrn 
approximately, and hence 



TORSION 147 

approximately. 

The resilience W of the spring is equal to one half the product of 
the force P multiplied by the axial extension or compression of the 
spring. Hence -, p2 27 



Problem 203. A helical spring is composed of 20 turns of steel wire .258 in. in 
diameter, the diameter of the coil being 3 in. If the spring is compressed by a 
force of 50 lb., what is the maximum stress in the spring, its axial compression, and 
its resilience ? 

108. General theory of spiral springs. The general theory of the 
cylindric spiral spring subjected to both axial load and torque has 
many important applications in physics and engineering, as, for ex- 
ample, in the helical-spring transmission dynamometer now coming 
into general use by reason of its ability to measure power without 
absorbing it. 

To analyze the most general case, suppose that an axial load P is 
applied to the spring and also a torque M t) the positive direction of P 
being chosen as that which will produce elongation of the spring, and 
the positive direction of M t as that which will increase the number 
of coils. Also let 

x = axial elongation of spring, 

cf) = angular rotation of spring, 

r = radius of coils, i.e. distance from center of wire to axis, 

a == angle of spiral (pitch angle), 

/ = length of wire, 

torque 
A = torsional rigidity = 



B = flexural rigidity = 



angle of twist per unit length 
bending moment 



flexure per unit length of wire- 
Note that if E= Young's modulus and G = shear modulus (modulus 
of rigidity), then B = El, where / denotes the static moment of inertia 
of a cross section of the wire with respect to its neutral axis, but that 
A is not equal to GI p , where I p denotes the polar moment of inertia of a 
cross section of the wire. The true values of A for various cross sec- 
tions, however, may be found from Articles 105-108 in connection 
with the above definition, and are summarized in the following table : 



148 



STRENGTH OF MATERIALS 



SHAIM: 


TORSION AL RIGIDITY 
A 


FLEXURAL RIGIDITY 
/^ 


r~n 


T TTGd* 


TTd^ 


vlx 1 


6, 32 


64 


^ 1 


Eq. 57, Art. 98 




1 i "" 


*-(-*> 

Eq. 58, Art. 98 


B = ^(J>_<Z*) 

64 


/^\" 


t ?r^ D 3 d 3 






16 D 2 + d 2 
Art. 103 


64 


e'~ 
n 


7T(? DM 3 


iJ DcZ 3 




16 D 2 +d 2 
Art. 103 


64 
















Gh* 






A 


"12 


, 


Art. 104 




-i 












1 1 

u 7-- -I -> I/, 


G b*h s 


^/i6 3 


' 

i 


3.5762+ / 4 2 


12 


6- 








Art. 104 




< _,,_ ;' 


G W 




i 

1 
1 

U-6--M " 


3.57 62 + / t 2 
Art. 104 


12 



TORSION 



149 






Now consider any cross section GOF of the wire (Fig. 105) and 
draw the F-axis through parallel to the axis of the helix, and the 
X-axis at right angles to OF and 
tangent to the cylinder on which 
the helix is wound. Also draw 
another pair of rectangular axes 
in the XOY- plane, namely OF 
tangent to the helix and OU nor- - 
mal to it. 

The axial load P produces a 
moment M b about OX of amount 
M b = Pr, while the torque M t acts 
about the axis OF. Represent 
each of these moments by a vec- 
tor, that is, a single line, the length of which represents the numerical 
amount of the moment, and the direction of which is the same as 
that in which a right-handed screw would advance if revolved in 

the direction indicated by the 





u X 



given moment. In the present 
case the torque M t causes rev- 
olution about the F-axis and 
is therefore represented by a 
vector laid off along this axis 
and pointing upward; while 
similarly the axial load P pro- 
duces revolution about the X- 
axis and is represented by a 
vector M b laid off along OX and 
pointing to the left (Fig. 106). 
Now in order to obtain the bending and twisting moments acting 

on the wire, these vectors must be resolved in the directions U and 

OF. Hence 

Moment about V (torsional moment) = M t sin a; -+- M b cos a, 
Moment about U (bending moment) = M t cos a M b sin a. 

Consequently, from the above definitions of torsional and flexural 
rigidity, we have 



FIG. 106 



150 



STRENGTH OF MATERIALS 



ATI...... Jf, sin a + M h cos a 

Angle of twist per unit of length = - 



Flexure per unit of length 



A 
M t cos a M h sin a 



To obtain the axial deflection of the spring and its angular rotation 
about its axis, these quantities must next be projected back on the 
X- and Y- axes. Making this projection (Fig. 107), we have 




FIG. 107 

Rotation about vertical axis Y per unit length of wire 

M f cos a M h sin a M t sin a -\- M h cos a . 

= - cosoH sin a, 

-t> A 

Rotation about horizontal axis OX per unit length of wire 



M sin a + M cos a 



cos a 



M cos a M sin a . 



B 



sin a. 



Multiplying each of these expressions by the length of the wire /, and 
simplifying, we have finally 



x = M h rl 



cos a sin a 
jj A 

1 



TOKSION 151 

Special Case I. Spirals very flat. 

In case the spirals are very flat, a may be as sumed to 1 e zero, and 
the above expressions then reduce to 

MJ, _ 

B 

Special Case II. Ifotatioii of ends prevented, i.e. (/> = 0. 

In this case first find M t in terms of M b from the equation < = ; 
then substitute this value of M t in the x equation and find x in terms 
of Jf 6 . 

Problem 204. Assume = and a = 45. Find x. 

Solution. Substituting these numerical values of and a in the expression 

/I 1\ /cos 2 a sin 2 

= M b l sin a cos al - - J + M t l ( 

and solving the resulting equation for M t , we have 



Substituting this value of M t in the equation 

,/cos 2 <* sin 2 -\ /I 1\ 

= M b rl / -- 1 -- J + M t lr sin a cos a / - - j, 

2 



x 
we have finally 



A + 



Special Case III. Axial elongation prevented, i.e. x = 0. 

This case applies to the helical-spring transmission dynamometer 
mentioned above. In this case first find M b in terms of M t from the 
equation x = 0, and then substitute this value of M b in the <f> equation 
and find < in terms of M t . 

Problem 205. A helical-spring transmission dynamometer is made of 15f turns 
of -in. steel wire, the mean diameter of the coil being 1^ in., and there being 2 turns 
of wire per inch. Calculate the torque required to produce an angular deflection 
of 25. 

Solution. From the table the constants A and B for a circular cross section are 

A = and B = Inserting these values in the equation x = 0, and assum- 
ing the relation between the elastic moduli to be G = f E, the result is 

sin a cos a 



A . , 
5 cos 2 a + 4 snr a 



152 STRENGTH OF MATERIALS 

From the given dimensions it is found that a = 7 15.4', and consequently 

M b = .02513 M t . 

Inserting this value in the <f> equation and also making d = .25, = , I = 62.35, 

and E = 30,000,000, the result of solving for M t is 

M t = 40.06 in. Ib. 



EXERCISES ON CHAPTER VI 

Problem 206. Find the diameter of a structural-steel engine shaft to transmit 
900 H.P. at 75 R.P.M. with a factor of safety of 10. 

Problem 207. Find the factor of safety for a wrought-iron shaft 5 in. in 
diameter which is transmitting 60 H.P. at 125 K.P.M. 

Problem 208. A structural-steel shaft is 60ft. long and is required to transmit 
500 H.P. with a factor of safety of 8 and to be of sufficient stiffness so that the 
angle of torsion shall not exceed .5 per foot of length. Find its diameter. 

Problem 209. Under the same conditions as in Problem 208 find the size of 
a hollow shaft if the external diameter is twice the internal. 

Problem 210. A hollow wrought-iron shaft 9 in. in external diameter and 
2 in. thick is required to transmit 600 H.P. with a factor of safety of 10. At what 
speed should it be run ? 

Problem 211. A horizontal steel shaft 4 in. in diameter and 10 ft. long be- 
tween centers of bearings carries a pulley weighing 300 Ib. and 14 in. in diameter. 
The belt on the pulley has a tension of 50 Ib. on the slack side and 175 Ib. on 
the driving side. Find the maximum combined stress in the shaft. 

Problem 212. An overhung steel crank carries a maximum thrust on the crank 
pin of 2 tons. Length of crank 9 in. ; distance from center of pin to center of 
bearing 5 in. Determine the size of crank and shaft for a factor of safety of 5. 

Problem 213. A propeller shaft 9 in. in diameter transmits 1000 H.P. at 90 
R.P.M. If the thrust on the screw is 12 tons, determine the maximum stress in 
the shaft. 

Problem 214. A round steel bar 2 in. in diameter, supported at points 4 ft. 
apart, deflects .029 in. under a central load of 300 Ib. and twists 1.62 in a length 
of 2 ft. under a twisting moment of 1500 ft. Ib. Find E, (?, and Poisson's ratio 
for the material (see Article 35). 

Problem 215. If P and Q denote the unit stresses at the elastic limits of a 

p 
material in tension and shear respectively, show that when <1 the material will 

P Q 

fail in tension, whereas when > 1 it will fail in shear, when subjected to com- 
bined bending and torsion, irrespective of the relative values of the bending and 
twisting moments. 

Solution. Combining Rankine's and Guest's formulas, we have 

16 M h 
P V = ' 



TORSION 153 

Consequently, if the bending moment is zero, p' = q' or = 1, whereas if it is 

<l' tf 
not zero, p' > q'. Similarly, if the twisting moment is zero, = 2. 

9' 
Now let F t and F s denote the factors of safety in tension and shear respectively. 

Then 

P 

Ft = = Ptf 
F s Q Qp>' 
Q' 

Since p' ==<?', the fraction = 1. Consequently, if <1 also, then <l;i.e. 
p' Q F s 

F t < F s and the material is weaker in tension than in shear. The second part of the 
theorem is proved in a similar manner. 

For a complete discussion of this question see article by A. L. Jenkins, En- 
gineering News (London, November 12, 1909), pp. 637-639. 

Problem 216. A steel -shaft subjected to combined bending and torsion has 
an elastic limit in tension of 64,600 lb./in. 2 and an elastic limit in shear of 
29,170 lb./in. 2 Show that Guest's formula applies to this material rather than 
Rankine's. 

Problem 217. A shaft subjected to combined bending and twisting is made of 
steel for which the elastic limit in tension is 28,800 lb./in. 2 and the elastic limit in 
shear is 16,000 lb./in. 2 Show that if the bending moment is one half the twisting 
moment, the shaft will be weakest in shear, whereas if the bending moment is twice 
the twisting moment, it will be weakest in tension. 

Problem 218. A closely coiled helical spring is made of 5-111. round steel wire 
and has 15 coils of mean diameter 3 in. Find its deflection under an axial 
.load of 20 lb., and the stiffness of the spring in pounds per foot of deflection. 

Problem 219. A helical-spring transmission dynamometer is made of 20 turns 
of J-in. steel wire ; mean diameter of coil 2| in. with 2 turns per inch. Find its 
axial twist in degrees when transmitting 6 H.P. at 75 R.P.M. 

Problem 220. A closely coiled helical spring is made of f-in. steel wire with 
coils of 3 in. mean diameter. Find the length of wire required in order that the 
spring shall deflect f-in. per pound of load. 



CHAPTER VII 

SPHERES AND CYLINDERS UNDER UNIFORM PRESSURE 

109. Hoop stress. When a hollow sphere or cylinder is subjected 
to uniform pressure, as in the case of steam boilers, standpipes, gas, 
water, and steani pipes, fire tubes, etc., the effect of the radial pres- 
sure is to produce stress in a circumferential direction, called hoop 
stress. In the case of a cylinder closed at the ends, the pressure on 
the ends produces longitudinal stress in the side walls in addition to 
the hoop stress. 

If the thickness of a cylinder or sphere is small as compared with 
its diameter, it is called a shell In analyzing the stress in a thin 
shell subjected to uniform pressure, such as 
that due to water, steam, or gas, it may be 
assumed that the hoop stress is distributed 
uniformly over any cross section of the shell. 
This assumption will be made in what follows. 
110. Hoop tension in hollow sphere. Con- 
sider a spherical shell subjected to^ uniform 
internal pressure, and suppose that the shell 

is cut into hemispheres by a diametral plane (Fig. 108). Then, if w 
denotes the pressure per unit of area within the shell, the resultant 

force acting on either hemisphere is P = > where d is the radius 

of the sphere. If p denotes the unit tensile stress on the circular 
cross section of the shell, the total stress on this cross section is Trdhp, 
approximately, where h is the thickness of the shell. Consequently, 

ird^w wd 

= irdlip ; whence p = - > 
4 4h 

which gives the hoop tension in terms of the radial pressure. 

From symmetry, the stress is the same on any diametral cross 
section. Therefore the equivalent stress at any point of the shell is 

154 




SPHERES AND CYLINDERS 



155 



m 1 
10 = - v 



l wd 
4/i, 



m m 

If the value of m is assumed to be 3^, this expression for p e becomes 

Problem 221. How great is the stress in a copper sphere 2 ft. in diameter and 
.25 of an inch thick, under an internal pressure of 175 lb./in. 2 ? 

111. Hoop tension in hollow circular cylinder. In the case of a 
cylindrical shell, its ends hold the cylindrical part together in such 
a way as to relieve the hoop tension at either extremity. Suppose, 
then, that the portion of the cylinder considered is so far removed 
from either end that the influence of the 
end constraint can be assumed to be zero. 

Suppose the cylinder cut in two by a 
plane through its axis, and consider a sec- 
tion cut out of either half cylinder by two 
planes perpendicular to the axis, at a dis- 
tance apart equal to c (Fig. 109). Then the 
resultant internal pressure P on the strip 
under consideration is P = cdw, and the resultant hoop tension is 
2 clip, where the letters have the same meaning as in the preceding 
article. Consequently, cdw = 2 clip ; whence 

div 




FIG. 109 



(64) 



p = 



If the longitudinal stress is zero, p e = p. 

This result is applicable to shells under both inner and outer pres- 
sure, if P is taken to be the excess of the internal over the external 
pressure. 

Problem 222. A cast-iron water pipe is 24 in. in diameter and 2 in. thick. 
What is the greatest internal pressure which it can withstand ? 

112. Longitudinal stress in hollow circular cylinder. If the ends 
of a cylinder are fastened to the* cylindrical part, the internal pres- 
sure against the ends produces longitudinal stresses in the side walls. 
In this case the cylindrical part is subjected both to hoop tension 
and to longitudinal tension. 



156 



STRENGTH OF MATERIALS 



To find the amount of the longitudinal tension, consider a cross 
section of the cylinder near its center, where the influence of the end 
restraints can be assumed to be zero (Fig. 110). Then the resultant 



\A 



ird'' 



W 





\ 


P 


P , 




1 > 







I 

IB 

FIG. 110 



pressure on either end is P = 

- 

and the resultant longitudinal stress on 
the cross section is Trdhp. Therefore 

ird\v wd 

= Trdhp ; whence p = 

4 4/& 

This is the same formula as for the 
sphere, which was to be expected, 
since the cross section is the same in both cases. 

If p l denotes the longitudinal stress and p h the hoop tension, then 

Pi = , p h = ; and, consequently, the equivalent stress p e is 



4 ill 



2 m 1 wd 



If m = 3^, this becomes 
(65) 





FIG. Ill 



Formula (65) is the one to be used in finding 
the tensile stress in a thin cylinder subjected to 
uniform internal pressure, in which the ends are 
held together by the body of 
the cylinder and not by inde- 
pendent stays or fixed sup- 
ports. 

Problem 223. An elevated water 
tank is cylindrical in form with 
a hemispherical bottom (Fig. 111). 
The diameter of the tank is 20 ft. 
and its height 52 ft., exclusive of 
the bottom. If the tank is to be 
built of wrought iron and the fac- 
tor of safety is taken to be 6, what 
should be the thickness of the bot- 
tom plates, and also of those in the 
body of the tank near its bottom ? FIG. 112 




NOTE. Formulas (63) and (65) give the required thickness of the plates, provided 
the tank is without joints. The bearing power of the rivets at the joints, however, is, in 
general, the consideration which determines the thickness of the plates (Art. 122). 



SPHERES AND CYLINDERS 



157 



B 



Problem 224 . A marine boiler shell is 16 ft. long, 8 ft. in diameter, and 1 in. thick. 
What is the stress in the shell for a working gauge pressure of 160 lb./in. 2 ? 

Problem 225. The air chamber of a pump is made of cast iron of the form 
shown in Fig. 112. If the diameter of the air chamber is 10 in. and its height 24 in., 
how thick must the walls of the air chamber be made to stand a pressure of 
500 lb./in. 2 with a factor of safety of 4? 

* 113. Differential equation of elastic curve for circular cylinder. 

A cyliudrical shell subjected to internal pressure is in a condition of 
stable equilibrium, for the internal pressure tends to preserve the 
cylindrical form of the shell, or to restore it to this form if, by any 
cause, the cylinder is flattened or otherwise deformed. A cylindrical 
shell which is subjected to external pressure, however, is in a con- 
dition of unstable equilibrium, for any deviation from a cylindrical 
form tends to be increased rather than 
diminished by the stress. In this respect 
thin hollow cylinders under external 
pressure are in a state of strain similar 
to that in a column, and the method of 
finding the critical pressure just preced- 
ing collapse is similar to that for finding 
the critical load for a column, as explained 
in the derivation of Euler's formula. 

Consider a thin hollow cylinder which 
is subjected to a uniform external pres- 
sure of amount w per unit of area, and suppose that in some way the 
cylinder has been compressed in one direction so that it assumes the 
flattened form shown in Fig. 113. The first step in the solution of 
the problem is to find the differential equation of the elastic curve in 
curvilinear coordinates, or, in other words, the differential equation 
of the elastic curve of the flattened cylinder referred to its original 
circular form. 

In polar coordinates let be the origin and OA the initial line. 
Also, let a denote the radius of the circular cylinder, and r the radius 
vector of the flattened or elliptical form. Now suppose that the cir- 
cular wall of the cylinder is considered as a piece which was origi- 
nally straight and has been made to assume a circular form by a 




FIG. 113 



* For a brief course the remainder of this chapter may be omitted. 



158 STRENGTH OF MATERIALS 

bending moment M'. Then, if p denotes the radius of curvature, from 
Article 66, - , 



Again, suppose that this circular cylinder is made to assume the 
flattened form as the result of an additional bending moment M, and 
let p f denote the corresponding radius of curvature. Then 

1 _ M' + M 
J'~ El 
Consequently, 

(ee) J~J = Jr 

From the differential calculus, 

2-ii 



dr\* 
) 
da da 2 



dr 

If the deformation is small, -- is infinitesimal, and r differs mfinitesi- 

da 

mally from a. Therefore, neglecting infinitesimals of an order higher 
than the second, the expression for p' becomes 



P' 



d 2 r d 2 r 

a 2 - a a-- 
da 2 da 2 



and, consequently, -, -, 1/72. 
p' a a? 

Since p = ,- = -, and therefore 
p a 

(67) ^~~p = ~ 

Comparing equations (66) and (67) 

( 68 > V^ =d 

^ / S-*4 sJ STS& 



*The sign is used because the calculus expression for p' contains a square root in 
the numerator. 



SPHERES AND CYLINDERS 



159 



Now let u denote the distanca between the circle and the ellipse 

measured radially. Then 

r = a u, 

or, if u is assumed to be positive when it lies outside the circle and 

negative when it lies inside, 

r = u + a. 

Differentiating both sides of this equation with respect to a, 



dr du 
da da 



d 2 r 



Also, if dl is the length of an infinitesimal arc of the circle, ada = dl. 
Substituting these values in equation (68), it becomes 



(69) 



which is the required differential equation of the elastic curve in the 
curvilinear coordinates I and u. 

114. Crushing strength of hollow circular cylinder. 'As a continu- 
ation of the preceding article, let it be required 
to find the external pressure which is just suffi- 
cient to cause the cylinder to retain its flattened 
form, or, in other words, the critical external pres- 
sure just preceding collapse. 

In Fig. 114 let OA and OB be axes of symmetry; 
then it is sufficient to consider merely the quadrant 
A OB. Let c denote the length of the chord AC, 
and let w be the unit external pressure. Then for 
a section of the cylinder of unit length the external pressure P on 

the curved strip A C is 

P = we. 

Now let M denote the bending moment at the point A. The tangen- 
tial force at this point is equal to the resultant pressure on OA, or wb. 
Consequently the bending moment M at the point C is 




114 



160 STRENGTH OF MATERIALS 

In the triangle OAC, 

~OC* = 'AC 1 + AC? -2AO- AD, or 

r 2 = c 2 + 6 2 - 2 b A D, 
from which 2 2 2 






Since r = u + a and a = b U Q) * 

in 

M= M Q + - (a 2 + 2 aw + ^ - a 2 - 2 



Since w and U Q are both infinitesimal, w c + u (or the difference between 
the absolute values of u and u ) is negligible in comparison with 2 . 
Therefore M= M - wa(u - u ), 

and, consequently, the differential equation of the elastic curve becomes 



The general integral of this differential equation is found to be 
(70) M = Wo + ^ + C lS i 

in which C l and (7 2 are the undetermined constants of integration.! 
This may be verified by substituting the integral in the above differ- 
ential equation. 

To determine C l and C 2 it is only necessary to make use of the 

terminal conditions at A and B. At the point A y I 0, = 0, and 

CLL 

u = u . Substituting these values in equation (70) and its first deriv- 
ative, it is found that 

a = and 0,,=-^- 

wa 

* Throughout this discussion it should be borne in mind that u is a negative quantity. 
t See Johnson, Treatise on Ordinary and Partial Differential Equations, 3d ed., 
pp. 85-86. 



SPHERES AND CYLINDERS 

Hence the integral becomes 

u = ^ + u -^cos l^l 
wa wa ^llI 

or 




At the upper end of the quadrant B the conditions are I = and 

= 0. Substituting these values in the first differential coefficient 
at 

obtained from equation (71), namely, 

I~H 
we have 



du M Iwa . Iwa 




whence 

Iwa 



EI 2 = " 

where X is an arbitrary integer. Choosing the smallest value of X, 
namely 1, this condition becomes 




wa a 
= 



whence 



If the thickness of the tube is denoted by h, then, for a section of 

h 8 

unit length, / = > and formula (72) becomes 
12 

(73) 

Formula (73) gives the critical pressure just preceding collapse ; that 
is to say, it gives the maximum external pressure w per unit of area 
which a cylindrical tube of thickness h can stand without crushing. 

Problem 226. What is the maximum external pressure which a cast-iron pipe 
18 in. in diameter and in. thick can stand without crushing ? 



162 



STEENGTH OF MATERIALS 



Problem 227. In a fire-tube boiler the tubes are of drawn steel, 2 in. internal 
diameter and | in. thick. What is the factor of safety for a working gauge pres- 
sure of 200 lb./in.2 ? 

115. Thick cylinders ; Lame's formulas. Consider a thick circular 
cylinder of external radius a and internal radius &, which is subjected 
to the action of either internal or external uniform pressure, or to 
both. Suppose a section is cut out of the cylinder by two planes per- 
pendicular to the axis at a unit's distance apart, and consider a small 
sector ABCD of angle a cut out of the ring so obtained, as shown 
in Fig. 115. Let p h denote the tangential stress, or hoop stress, acting on 
tin's infinitesimal element, p r the radial stress acting on the inner sur- 
face AD, and p r + dp r the radial stress acting on the outer surface BC. 




FIG. 115 

Then the internal and external radii being r and r + dr respectively, 
the length of AD is ra and of BC is (r -f dr)a. Since the width 
of the piece is unity, the resultant radial force acting on the piece, or 
the difference between the pressure on the inner and outer surfaces, 
is (p r + dp r ) (r -f dr) a p r ra. Therefore, since the resultant of the 
hoop stress in a radial direction is (p h a) dr, in order that the radial 
stresses shall equilibrate, 

(p r + dp r ) (r + dr) a p r ra = p h adr ; 
or, neglecting infinitesimals of an order higher than the second, 

p r dr + rdp r = p h dr ; 
which may be written 

(74) - 



SPHERES AND CYLINDERS 163 

If the ends of the cylinder are free from restraint, or if the cylinder 
is subjected to a uniform longitudinal stress, the longitudinal defor- 
mation must be constant throughout the cylinder. The longitudinal 
deformation, however, is due to the lateral action of p r and p h , and is 

7) 1) 1 

of amount -^- -f -~- > or - ( p r + p h ), in which m denotes Poisson's 
mE mE mE 

constant. Therefore, if this expression is constant, p r + p h must 

be constant, and hence _ -, 

Pr ~r Ph K > 

where k is a constant. Consequently, p h = k p r , and substituting 
this value of p h in equation (74) and multiplying by r, it becomes 

krdr = 2 rp r dr -\- r 1 dp r , 
which may be written ^ 

dr 
Integrating, ^ 

in which C l is the constant of integration ; whence 
Also, since p h = k p r , 



Now suppose that the cylinder is subjected to a uniform internal 

pressure of amount w i per unit of area, and also to a uniform external 

pressure of amount w e per unit of area. Then p r w e when r a, 

and p r = w i when r = 1. Substituting these values in equation (75), 

i- r 1 i- r^ 

/v V/i tu \j *- 

whence 2 7 2 / _ \ o 



Therefore, substituting these values of <7 X and k in equations (75) and 

(76), they become 

_ w e a z Wjb 2 a z b 2 (w e 

(77) 

. a'Q'( w e - 

I /^.2 i,2\ 



164 STRENGTH OF MATERIALS 

which give the radial and hoop stresses in a thick cylinder subjected 
to internal and external pressure. Equations (77) are known as 
Lamp's formulas. 

116. Maximum stress in thick cylinder under uniform internal 
pressure. Consider a thick circular cylinder which is subjected only 
to internal pressure. Then w e = 0, and equations (77) become 

w a? \ w / 2 

(78) 



Since p h is negative, the hoop stress in this case is tension. 

Since p r and p h both increase as r decreases, the maximum stress 
occurs on the inner surface of the cylinder, where 

w-(a -\- b ) 
T = 0. f) = w., and t), = 

JJ T 1* M ft 2 Z.2 

Clearing the latter of fractions, we have = l - , whence the 

thickness of the tube, h = a &, is given by 

(79) h = b\ 



Moreover, the equivalent stress for a point on the inner surface of 
the cylinder is 



If m = 3 J, the absolute value of the equivalent stress becomes 



This may also be written 



Problem 228. Find the thickness necessary to give to a steel locomotive cylinder 
of 22 in. internal diameter, if it is required to withstand a maximum steam pressure 
of 150 lb./in. 2 with a factor of safety of 10. 

Problem 229. In a four-cycle gas engine the cylinder is of steel with an internal 
diameter of 6 in., and the initial internal pressure is 200 lb./in. 2 absolute. With 
a factor of safety of 15, how thick should the walls of the cylinder be made ? 

Problem 230. The steel cylinder of an hydraulic press has an internal diameter 
of 5 in. and an external diameter of 7 in. With a factor of safety of 3, how great 
an internal pressure can the cylinder withstand ? 



SPHEEES AND CYLINDERS 165 

117. Bursting pressure for thick cylinder. Let ^denote the ulti- 
mate tensile strength of the material of which the cylinder is com- 
posed. Then, from equation (79), the maximum allowable internal 
pressure w i is obtained from the equation 



whence 



If m is assumed to be 3J, this formula becomes 

u t (a* - ft 2 ) 



Equations (81) and (82) give the maximum internal pressure w { 
which the cylinder can stand without bursting. 

Problem 231. A wrought-iron pipe is 4 in. in external diameter and .25 in. 
thick. What head of water will it stand without bursting? 

Problem 232. Under a water head of 200 ft., what is the factor of safety in 
the preceding problem ? 

118. Maximum stress in thick cylinder under uniform external 
pressure. Consider a thick circular cylinder subjected only to external 
pressure. In this case w i = and equations (77) become 



Since p h is positive, the hoop stress in this case is compression. 
For a point on the inner surface of the cylinder 

2 w e a 2 
r = b, p r = Q, and P h = ^p' 

Since the radial stress is zero on the inner surface, the equivalent 
stress is equal to the hoop stress, that is, 



Problem 233. A wrought-iron cylinder is 8 in. in external diameter and \\ in. 
thick. How great an external pressure can it withstand ? 



166 STRENGTH OF MATERIALS 

119. Thick cylinders built up of concentric tubes. From equa- 
tions (77), it is evident that in a thick cylinder subjected to internal 
pressure the stress is greatest on the inside of the cylinder, and 
decreases toward the outside. In order to equalize the stress through- 
out the cylinder and thus obtain a more economical use of material, 
the device is resorted to of forming the cylinder of several concentric 
tubes and producing an initial compressive stress on the inner ones. 
For instance, in constructing the barrel of a cannon, or the cylinder 
of an hydraulic press, the cylinder is built up of two or more tubes. 
The outer tubes in this case are made of somewhat smaller diameter 
than the inner tubes, and then each is heated until it has expanded 
sufficiently to be slipped over the one next smaller. In cooling, the 
metal of the outer tube contracts, thus producing a compressive stress 
in the inner tube and a tensile stress in the outer tube. If, then, this 
composite tube is subjected to internal pressure, the first effect of 
the hoop tension thus produced is to relieve the initial compressive 
stress in the inner tube and increase that in the outer tube. Thus 
the resultant stress in the inner tube is equal to the difference between 
the initial stress and that due to the internal pressure, whereas the 
resultant stress in the outer tube is the sum of these two. In this 
way the strain is distributed more equally throughout the cylinder. 
It is evident that the greater the number of tubes used in building 
up the cylinder, the more nearly can the strain be equalized. 

The preceding discussion of the stress in thick tubes can also be 
applied to the calculation of the stress in a rotating disk. For example, 
a grindstone is strained in precisely the same way as a thick tube 
under internal pressure, the load in this case being due to centrifugal 
force instead of to the pressure of a fluid or gas. 

120. Practical formulas for the collapse of tubes under external 
pressure. A more rigorous analysis of the stress in thin tubes, due 
to external pressure, than that given in Article 114, using Poisson's 

ratio of transverse to longitudinal deformation, gives the formula * 
m 

E 

w 



4 l-- 2 

v \ 

* Love, Math. Theory ofElast., Vol. II, pp. 308-316. 



SPHEKES AND CYLINDERS 167 

or, m terms of the diameter D = 2 a, 




This formula, however, is based on the assumptions that the tube is 
perfectly symmetrical, of uniform thickness, and of homogeneous 
material, conditions which are never fully realized in commercial 
tubes. From recent experiments 011 the collapse of tubes,* however, 
it is now possible to determine the practical limitations of this for- 
mula, and so modify it, by a method similar to that by which the 
Gordon-Eankine column formula was deduced from Euler's formula 
(Articles 88, 89), as to obtain a rational formula which shall never- 
theless conform closely to experimental results. By determining the 
ellipticity, or deviation from roundness, and the variation in thick- 
ness of the various types of tubes covered by the tests mentioned 
above, it is found that by introducing empirical constants the rational 
formulas can be made to fit experimental results as closely as any 
empirical formulas, with the advantage of being unlimited in their 
range of application.! The formula so obtained is 

ffor thin tubes 

t 5 

where h = average thickness of tube in inches, 

D = maximum outside diameter in inches, 

= Poisson's ratio = .3 for steel, 
m 

C = .69 for lap-welded steel boiler flues, 
= .76 for cold-drawn seamless steel flues, 
= .78 for drawn seamless brass tubes. 

By a similar procedure for thick tubes / >.023) a practical 

* Carman, " Resist, of Tubes to Collapse," Univ. HI. Bull., Vol. Ill, No. 17 ; Stewart, 
"Collap. Press. Lap-Welded Steel Tubes," Trans. A.S.M.E., 1906, pp. 730-820. 

t Slocum, " The Collapse of Tubes under External Pressure," Engineering. London, 
January 8, 1909. Also abstract of same article in Kent, 8th ed., 1910, pp. 320-322. 




168 



STRENGTH OF MATERIALS 



rational formula has been obtained from Lame's formula, Article 118, 
for this case also, namely 

IK) 



w 



D 



for thick tubes 



where 



u c = ultimate compressive strength of the material, 
K = .89 for lap- welded steel boiler flues. 

Only one value of K is given, as the experiments cited were all made 
on one type of tube. 

The correction constants C and K include corrections both for 
ellipticity, or flattening of the tube, and for variation in thickness. 
Thus if the correction for ellipticity is denoted by C l and the correc- 
tion for variation in thickness by <7 2 , we have 

Minimum outside diameter 
1 



_ 



Maximum outside diameter 
Minimum thickness 



Average thickness 
and the correction constants C and K are therefore denned as 



By an "experimental determination of C 1 and C 2 the formulas can 

therefore be applied to any given type of tube. 

121. Shrinkage and forced fits. In machine construction shrink- 

age and forced or pressed fits are frequently employed for connecting 

certain parts, such as crank disk and shaft, wheel and axle, etc. To 

make such a connection the 
shaft is finished slightly larger 
than the hole in the disk or 
ring in which it belongs. The 
shaft is then either tapered 
slightly at the end and pressed 
into the ring cold, or the ring 
is enlarged by heating until it 

will slip over the shaft, in which case the shrinkage due to cooling 

causes it to grip the shaft, 



Di 



D 



FlG 



SPHERES AND CYLINDERS 169 

To analyze the stresses arising from shrinkage and forced fits, let 
D l denote the diameter of the hole in the ring or disk, and Z> 2 the 
diameter of the shaft (Fig. 116). When shrunk or forced together, 
D l must increase slightly and D 2 decrease slightly, i.e. D l and Z> 2 
must of necessity take the same value D. Consequently the circum- 
ference of the hole changes from 7rD l to 7rZ>, and hence the unit 
deformation s x of a fiber on the inner surface of the hole is 



l 7TD, D 

Similarly the unit deformation s 2 of a fiber on the surface of the 
Shaftis 



A 

From Hooke's law, = E y we have therefore for the unit stress p l 
s 

on the inside of the disk 

i = . 1 = :^; 

E, D, 

and for the unit stress p on the surface of the shaft 



Adding these two equations to eliminate the unknown quantity 
the result is 



where K denotes the allowance, or difference in diameter of shaft and 
hole. For a thick disk or heavy ring this allowance K may be deter- 
mined from the nominal diameter D of the shaft by means of the 
following empirical formulas.* 



For shrinkage fits, K = \ , 

For pressed fits, K = ^ > 

For driven fits, K = ^ 2 

* S. H. Moore, Tram. Am. Soc. Mech. Eng., Vol. XXIV. 



170 STRENGTH OF MATERIALS 

For thin rings, however, the allowance given by these formulas will 
be found to produce stresses in the ring entirely too large for safety. 
In deciding on the allowance for any given class of work, the working 
stresses in shaft and ring may first be assigned and the allowance 
then determined from the formulas given below so that the actual 
stresses shall not exceed these values. 

From Lame's formulas the stresses p l and p 2 may be obtained in 
terms of the unit pressure between the surfaces in contact. Thus 
from formula (80) the equivalent stress on the inside of the hole is 

A =A = - (-7^ + 1.819, 



where Z> 3 denotes the outside diameter of the ring, while, by substi- 
tuting r = a and b = in the equations of Article 118, the stresses 
on the outer surface of the shaft are found to be 

Ph = W > Pr = > 

and consequently -, 

P2=Ph-Pr = -7w- 

Eliminating w between these expressions for p^ and p , we have 



~ .7 

Now to simplify the solution, let the coefficient of p 2 be denoted by 
H; that is, let 

U == > 

rr / T\2 7~)2\ 

in which case 

p l = Hp 2 . 

Eliminating p l between this relation and the above expression for the 
allowance K, we have finally 






Pi = Hp z . 

In applying these formulas the constant .If is first computed from the 
given dimensions qf the parts. If the allowance K is given, the unit 



SPHERES AND CYLINDERS 171 

stresses p l and p 2 in ring and shaft are then found from the above. 
If K is to be determined, a safe value for the stress in the ring, p , is 
assigned, and p 2 calculated from the second equation. This value is 
then substituted in the first equation and K calculated. 

The following problem illustrates the application of the formulas. 

Problem 234. A cast-iron gear, 8 in. external diameter, 3 in. wide, and If in. 
internal diameter, is to be forced on a steel shaft. Find the stresses developed, the 
pressure required to force the gear on the shaft, and the tangential thrust required 
to'shear the fit, i.e. produce relative motion between gear and shaft. 

Solution. From the formula K 2 the allowance is found to be .004 in., 

1000 

making the diameter of the shaft D 2 = 1.754 in. Also since DI = 1.75 in., D 3 = 8 in., 
we have H - 2.0007. Hence assuming EI = 15,000,000 lb./in. 2 and E 2 = 30,000,000 
lb./in. 2 , we have 

p l = 13,713 lb./in. 2 , p 2 = 27,436 lb./in. 2 

To find the pressure required to force the gear on the shaft it is first necessary 
to calculate the pressure between the surfaces in contact. From the relation 
p 2 = . 7 w this amounts to 

u> = 39,194 lb./in. 2 

The coefficient of friction depends on the nature of the surfaces in contact. As- 
suming it to be /t = .15 as an average value, and with a nominal area of contact 
of TT x If x 3 = 16. 485 in. 2 , the total pressure P required is 

P = 16.485 x 39,194 x .15 = 96,917 Ib. = 48.5 tons. 
To find the torsional resistance of the fit, we have, as above 

Bearing area = 16.485 in. 2 , Unit pressure = 39,194 lb./in. 2 , 

/x, = .15, radius of shaft = .875 in. 
Hence the torsional resistance is 

M t = 16.485 x 39,194 x .15 x .875 = 85,000 in. Ib. 

Consequently the tangential thrust on the teeth of the gear necessary to shear 
the fit is 

isoao _ 21,250 Ib. = 10.6 tons. 

122. Riveted joints. In structural work such as plate girders, 
trusses, etc., and also in steam boilers, standpipes, and similar con- 
structions, the connections between the various members are made 
by riveting the parts together. As the holes for the rivets weaken 
the members so joined, the strength of the structure is determined 
by the strength of the joint. 

Failure of a riveted joint may occur in various ways, namely, by shear- 
ing across the rivet, by crushing the rivet, by crushing the plate in 



172 STRENGTH OF MATERIALS 



front of the rivet, by shearing the plate, i.e. pulling out the rivets, or 
by tearing the plate along the line of rivet holes. Experience has 
shown, however, that failure usually occurs either by shearing across 
the rivet or by tearing the plate along the line of rivet holes. 

The strength of any given type of riveted joint is expressed by 
what is called its efficiency, denned as 

strength of joint 



Efficiency of riveted joint = 



strength of unriveted member 



Thus if d (Fig. 117), denotes the diameter of a rivet and c the distance 
between rivet holes, or pitch of the rivets as it is called, the efficiency 
of the joint against tearing of the plate along the line of rivets is 

c d 



e = 



To determine the efficiency of the joint against shearing across the 
rivets, let q denote the ultimate shearing strength of the rivet and p 
the ultimate tensile strength of the plate. Then for a single lap joint 
(Fig. 117), if h denotes the thickness of the plate, the area corre- 

J2 

sponding to one rivet is hd t and the area in shear for each rivet is ; 
consequently the efficiency of this type of joint against rivet shearing is 



c = 



For an economical design these two efficiencies should be equal. For 
practical reasons, however, it is not generally possible to make these 
exactly equal, and in this case the smaller of the two determines 
the strength of the joint. 

For a double-riveted lap joint the efficiency against tearing of the 

plate is 

c d 



as above ; but since in this case there are two rivets for each strip of 
length c, the efficiency against rivet shear is 



2chp 



SPHERES AND CYLINDERS 



173 




SINGLE-RIVETED LAP JOINT 
EFFICIENCY 50-60 PER CENT 



SINGLE-RIVETED BUTT JOINT 
EFFICIENCY 76-78 PER CENT 




DOUBLE-RIVETED LAP JOINT 
EFFICIENCY 70-72 PER CENT 



i*-c-*i 



> ( 

y) 





DOUBLE-RIVETED BUTT JOINT 
EFFICIENCY 82-83 PER CENT 



FIG. 117 



Similarly for a single-riveted butt joint with two cover plates the 
efficiency of the joint against tearing of the plate is 



c d 



and against rivet shear is 



e = 



2 chp 



For a double-riveted butt joint with two cover plates the efficiency 
against tearing of the plate is 

c d 



e = 



174 STRENGTH OF MATERIALS 

and against rivet shear is 



clip 

The average efficiencies of various types of riveted joints as used in 
steam boilers are given in Fig. 117. 

In designing steam-boiler shells it is customary in this country to 
determine first the thickness of shell plates by the following rule. 

To find the thickness of shell plates, multiply the maximum steam 
pressure to be carried (safe working pressure in lb./in. 2 ) by half the 
diameter of the boiler in inches. This gives the hoop stress in the 
shell per unit of length. Divide this result by the safe working stress 
(working stress = ultimate strength, usually about 60,000 lb./in. 2 , 
divided by the factor of safety, say 4 or 5) and divide the quotient 
by the average efficiency of the style of joint to be used, expressed as 
a decimal. The result will be the thickness of the shell plates ex- 
pressed in decimal fractions of an inch. 

Having determined the thickness of shell plates by this method, 
the diameter of the rivets is next found from the empirical formula 

d = k^ft, 

where k.= 1.5 for lap joints and Jc =1.3 for butt joints with two 
cover plates. 

The pitch of the rivets is next determined by equating the strength 
of the plate along a section through the rivet holes to the strength 
of the rivets in shear, and solving the resulting equation for c. 

To illustrate the application of these rules, let it be required to 
design a boiler shell 48 in. in diameter to carry a steam pressure of 
125 lb./in. 2 with a double-riveted, double-strapped butt joint. 

By the above rule for thickness of shell plates we have 

125 X -V- 

h = - 2 = .3, say -A m. 
e 0,000 x 82 

5 

The diameter of rivets is then 



d =1.3 = .73, say f in. 

To determine the pitch of the rivets, the strength of the plate for a 
section of width c on a line through the rivet holes is 

c- 5 X 60,000, 



SPHERES AND CYLINDERS 175 

and the strength of the rivets in shear for a strip of this width is 



4 x-q = 7r~x 40,000. 
4 J ID 

Equating these two results and solving for c, we have 

( c -!)i 5 6 x 60,000 = 71-^ x 40,000, 
whence 

c = 4. 5 in. 

As a check on the correctness of our assumptions the efficiency of 
the joint is found to be 



- 

c 4.5 

For bridge and structural work the following^ enr^Fical rules are 
representative of American practice.* 

The pitch or distance from center to center of rivets should not 
be less than 3 diameters of the rivet. In bridge work the pitch should 
not exceed 6 inches, or 1 6 times the thickness of the thinnest outside 
plates except in special cases hereafter noted. In the flanges of beams 
and girders, where plates more than 12 inches wide are used, an extra 
line of rivets with a pitch not greater than 9 inches should be driven 
along each edge to draw the plates together. 

At the ends of compression members the pitch should not exceed 4 
diameters of the rivet for a length equal to twice the width or diameter 
of the member. 

In the flanges of girders and chords carrying floors, the pitch should 
not exceed 4 inches. 

For plates in compression the pitch in the direction of the line of 
stress should not exceed 16 times the thickness of the plate, and 
the pitch in a direction at right angles to the line of stress should 
not exceed 32 times the thickness, except for cover plates of top 
chords and end posts in which the pitch should not exceed 40 times 
their thickness. 

The distance between the edge of any piece and the center of the 
rivet hole should not be less than 1J inches for |-inch and |-inch 
rivets except in bars less than 2J- inches wide ; when practicable it 

* Given by Cambria Steel Co. 



176 STRENGTH OF MATEEIALS 

should, for all sizes, be at least 2 diameters of the rivet and should 
not exceed 8 times the thickness of the plate. 

EXERCISES ON CHAPTER VII 

Problem 235. The end plates of a boiler shell are curved out to a radius of 5 ft. 
If the plates are | in. thick, find the tensile stress due to a steam pressure of 
175 lb./in. 2 

Problem 236. If the thickness of the end plates in Problem 235 is changed to 
\ in., the steam pressure being the same, to what radius should they be curved 
in order that the tensile stress in them shall remain the same ? 

Problem 237. In a double-riveted lap joint the plates are \ in. thick, rivets 
| in. in diameter, and pitch 3 in. Calculate the efficiency of the joint. 

Problem 238. A boiler shell is to be 4 ft. in diameter, with double-riveted lap 
joints, and is to carry a steam pressure of 90 lb./in. 2 with a factor of safety of 5. 
Determine the thickness of shell plates, and diameter and pitch of rivets. Also 
calculate the efficiency of the joint. 

Problem 239. A cylindrical standpipe is 75 ft. high and 25 ft. inside diameter, 
with double-riveted, two-strap butt joints. Determine the required thickness of 
plates near the bottom for a factor of safety of 5, and also the diameter and pitch 
of rivets. 

Problem 240. The cylinder of an hydraulic press is 12 in. in diameter. Find 
its thickness to stand a pressure of 1500 lb./in. 2 , if it is made of cast iron and the 
factor of safety is 10. 

Problem 241. A high-pressure, cast-iron water main is 4 in. inside diameter and 
carries a pressure of 800 lb./in. 2 Find its thickness for a factor of safety of 15. 

Problem 242. The water chamber of a fire engine has a spherical top 18 in. 
in diameter, and carries a pressure of 250 lb./in. 2 It is made of No. 7 B. and S. 
gauge copper, which is reduced in manufacture to a thickness of about .1 in. 
Determine the factor of safety. 

Problem 243. A cast-iron ring 3 in. thick and 8 in. wide is forced on a steel 
shaft 10 in. in diameter. Find the stresses in ring and shaft, the pressure required 
to force the ring on the shaft, and the torsional resistance of the fit. 

NOTE. Since the ring in this case is relatively thin, assume an allowance of about 
half the amount given by Moore's formula. Then having given Dz = 10 in., Ds = 13 in., 
and computed the allowance K, we have also D\ = Di K, and inserting these values 
in the formulas of Article 121, the required quantities may be found, as explained in 
Problem 234. 

Problem 244. The following data are taken from Stewart's experiments on the 
collapse of thin tubes under external pressure, the tubes used for experiment 
being lap-welded, steel boiler flues. Compute the collapsing pressure from the 
rational formula for thin tubes, given in Article 120, for both the average thickness 
and least thickness, and note that these two results lie on opposite sides of the 
value obtained directly by experiment. 



SPHERES AND CYLINDERS 



177 



OUTSIDE DIAMETER IN INCHES 


THICKNESS h IN INCHES 


ACTUAL 






COLLAPS- 




At place of collapse 




At place of collapse 


ING 


Average 






Average 




T> 




Greatest = D 


Least = d 




Greatest 


Least 


Ib./in.z 


8.604 


8.610 


8.580 


0.219 


0.230 


0.210 


870 


8.664 


8.670 


8.625 


0.226 


0.227 


0.204 


840 


8.665 


8.670 


8.660 


0.212 


0.240 


0.211 


880 


8.653 


8.665 


8.590 


0.208 


0.220 


0.203 


970 


8.688 


8.715 


8.605 


0.274 


0.280 


0.266 


1430 


8.664 


8.695 


8.635 


0.258 


0.261 


0.248 


1320 


8.645 


8.665 


8.635 


0.263 


0.268 


0.259 


1590 


8.674 


8.675 


8.675 


0.273 


0.282 


0.270 


2030 


8.638 


8.645 


8.615 


0.279 


0.298 


0.280 


2200 


10.055 


10.180 


9.950 


0.157 


0.182 


0.150 


210 



Problem 245. The following data are taken from Stewart's experiments on the 
collapse of thick tubes under external pressure. The ultimate compressive strength 
of the material was not given by the experimenter, but from the other elastic 
properties given, it is here assumed to be u c = 38,500 lb./in. 2 Compute the col- 
lapsing pressure from the rational formula for thick tubes, given in Article 120, 
for both average and least thickness, and compare these results with the actual 
collapsing pressure obtained by experiment. 



OUTSIDE DIAMETER IN INCHES 


THICKNESS h IN INCHES 


ACTUAL 






COLL AT s- 




At place of collapse 




At place of collapse 


INO 


Average 












Greatest = D 


Least = d 




Greatest 


Least 


lb./in.* 


4.010 


4.020 


3.980 


0.173 


0.203 


0.140 


2050 


4.014 


4.050 


3.990 


0.178 


0.277 


0.158 


2225 


4.012 


4.050 


3.960 


0.173 


0.200 


0.170 


2425 


4.018 


4.050 


4.010 


0.184 


0.192 


0.165 


2540 


2.997 


3.010 


2.980 


0.147 


0.151 


0.138 


3350 


2.987 


3.010 


2.970 


0.139 


0.139 


0.125 


2575 


2.990 


3.010 


2.970 


0.190 


0.218 


0.166 


4200 


2.996 


3.020 


2.980 


0.191 


0.216 


0.176 


4200 


2.997 


3.020 


2.960 


0.190 


0.215 


0.161 


4175 


3.000 


3.020 


2.960 


0.182 


0.192 


0.165 


3700 



Problem 246. A boiler shell f in. thick and 5 ft. in diameter has longitudinal, 
single-riveted lap joints, with 1-in. rivets and 2^-in. rivet pitch. Calculate the 
maximum steam pressure which can be used with a factor of safety of 5. 



178 STRENGTH OF MATERIALS 

Problem 247. A cylindrical standpipe 80 ft. high and 20 ft. inside diameter is 
made of ^-in. plates at the base with longitudinal, double-riveted, two-strap butt 
joints, connected by 1-in. rivets with a pitch of 3 in. Compute the factor of safety 
when the pipe is full of water. 

Problem 248. In a single-riveted lap joint calculate the pitch of the rivets and 
the distance from the center of the rivets to the edge of the plate under the assump- 
tion that the diameter of the rivet is twice as great as the thickness of the plate. 

Solution. Consider a strip of width equal to the rivet pitch, i.e. a strip contain- 
ing one rivet. Let q denote the shearing strength of the rivet, and p the tensile of 
the plate. Then if h denotes the thickness of the plate, in order that the shearing 
strength of the rivet may be equal to the tensile strength of the plate along the line 
of rivet holes, we must have 

TTd 2 

q = (c-d)hp. 

Since the rivet is usually of better material than the plate, we may assume that 
the ultimate shearing strength of the rivet is equal to the ultimate tensile strength 
of the plate, i.e. assume that p = q. Under this assumption the above relation 
becomes 



whence 

c = 2.5d, approximately. 

Similarly, in order that the joint may be equally secure against shearing off the 
rivet and pulling it out of the plate, i.e. shearing the plate in front of the rivet, 
the condition is 



where a denotes the "margin," or distance from center of rivets to edge of plate, 

' 4 

and q' denotes the ultimate shearing strength of the plate. Assuming that q' = -q 

d 5 

and h = - , and solving the resulting expression for a, we have 



CHAPTER VIII 

FLAT PLATES 

123. Theory of flat plates. The analysis of stress in flat plates is, 
at present, the most unsatisfactory part of the strength of materials. 
Although flat plates are of frequent occurrence in engineering con- 
structions, as, for example, in manhole covers, cylinder ends, floor 
panels, etc., no general theory of such plates has as yet been given. 
Each form of plate is treated by a special method, which, in most 
cases, is based upon an arbitrary assumption as to the dangerous 
section, or the reactions of the supports, and therefore leads to 
questionable results. 

Although the present theory of flat plates is plainly inadequate, 
it is, nevertheless, of value in pointing out the conditions to which 
such plates are subject, and furnishing a rational basis for the esti- 
mation of their strength. The formulas derived in the following 
paragraphs, if used in this way, with a clear understanding of their 
approximate nature, will be found to be invaluable in designing, or 
determining the strength of flat plates. 

The following has come to be the standard method of treatment, 
and is chiefly due to Bach.* 

124. Maximum stress in homogeneous circular plate under uni- 
form load. Consider a flat, circular plate of homogeneous material, 
which bears a uniform load of amount w per unit of area, and suppose 
that the edge of the plate rests freely on a circular rim slightly 
smaller than the plate, every point of the rim being maintained at 
the same level. The strain in this case is greater than if the plate 
was fixed at the edges, and, consequently, the formula deduced will 
give the maximum stress in all cases. 

* For an approximate method of solution see article by S. E. Slocum entitled "The 
Strength of Flat Plates, with an Application to Concrete-Steel Floor Panels," Engineer- 
ing News, July 7, 1904. 

179 



180 



STEENGTH OF MATERIALS 



Now suppose a diametral section of the plate taken, and regard 
either half of the plate as a cantilever (Fig. 118). Then if T is the 
radius of the plate, the total load on this semi- 

vrr 2 
circle is - w, and its resultant is applied at 

the center of gravity of the semicircle, which is 

4r 

at a distance of from AB. The moment of 
3?r 

this resultant about the support AB is therefore 

ITf* 4 IT 2i "7* iJ} 

-W- > or - Similarly, the resultant 

2 O 7T O 

of the supporting forces at the edge of the 
plate is of amount - -w, and is applied at the center of gravity of 

2r 
the semi-circumference, which is at a distance of - - from AB. The 

7T 




FIG. 118 



Trr 2 



moment of this resultant about AB is therefore 



TTT Z W 2 r 



> or r w. 



7T 



Hence the total external moment M at the support is 



Now assume that the stress at any point of the plate is independ- 
ent of the distance of this point from the center. Under this arbi- 
trary assumption the stress in the plate is given by the fundamental 
formula in the theory of beams, namely, 

Me 

P--J- 

If the thickness of the plate is denoted by h, then, since the breadth 
of the section is ~b = 2 r, 



Consequently, 



l>li* rh s h 

= To' = '""' and e = o' 

JLZ D 



Me 
- 



whence 
(83) 



FLAT PLATES 181 

Foppl has shown that the arbitrary assumption made in deriving 
this formula can be avoided, and the same result obtained, by a more 
rigorous analysis than the preceding; and Bach has verified the 
formula experimentally. Formula (83) is therefore well established 
both theoretically and practically. 

Problem 249. The cylinder of a locomotive is 20 in. internal diameter. What 
must be the thickness of the steel end plate if it is required to withstand a pres- 
sure of 160 lb./in. 2 with a factor of safety of 6 ? 

Problem 250. A circular cast-iron valve gate i in. thick closes an opening 6 in. 
in diameter. If the pressure against the gate is due to a water head of 150 ft., 
what is the maximum stress in the gate ? 

125. Maximum stress in homogeneous circular plate under con- 
centrated load. Consider a flat, circular plate of homogeneous mate- 
rial, and suppose that it bears a single concentrated load P which is 
distributed over a small circle of radius r Q concentric with the plate. 
Taking a section through the center of the plate and regarding either 
half as a cantilever, as in the preceding article, the total rim pres- 

P 2r 

sure is , and it is applied at a distance of from the center. The 

p 7T 

total load on the semicircle of radius r is > and it is applied at a dis- 

4r 
tance of - from the section. Therefore the total external moment M 

3-7T 



at the section is 



^ = ^_2Pn = p r / 1 _2nY 

7T 3-7T 7T \ 3r 



Assuming that the stress is uniformly distributed throughout the plate, 
the stress due to the external moment M is given by the formula 

Me 
P = -j' 

If the thickness of the plate is denoted by k, then 

rh s h 

/= _ and , = _. 

Therefore f 

Me ~i 



whence 



182 



STRENGTH OF MATERIALS 



If r = 0, that is to say, if the load is assumed to be concentrated 
at a single point at the center of the plate, formula (84) becomes 



(85) 



P = 



_ 

TTh 2 



If the load is uniformly distributed over the entire plate, then 
= r and P = Trr*w, where w is the load per unit of area. In this 



case formula (84) becomes 



o 



which agrees with the result of the preceding article. 

Problem 251. Show that the maximum concentrated load which can be borne 
by a circular plate is independent of the radius of the plate. 

126. Dangerous section of elliptical plate. Consider a homo- 
geneous elliptical plate of semi-axes a and b and thickness h, and 
G suppose that an axial cross is cut out 

of the plate, composed of two strips 
AB and CD, each of unit width, and 
intersecting in the center of the plate, 
as shown in Fig. 119. 

Now suppose that a single concen- 
trated load acts at the intersection 
of the cross and is distributed to the 
supports in such a way that the two 
beams AB and CD each deflect the same amount at the center. Since 
AB is of length 2 a, from Article 67, Problem 119, the deflection 

P C2 a} s 
at the center of AB is D l - ' From symmetry, the reactions 

at A and B are equal. Therefore, if each of these reactions is denoted 
by R 19 2 R^ = P, and, consequently, 




3 JET 

Similarly, if R z denotes the equal reactions at C and D, the deflec 
tion !> of CD at its center is 



FLAT PLATES 183 

If the plate remains intact, the two strips AB and CD must deflect 
the same amount at the center. Therefore D^ = D 2 , and hence 

1-5- 

For the beam AB of length 2 a the maximum external moment is 

^3 j 

R^a. Also, since AB is assumed to be of unit width, / = and e = - 
Hence the maximum stress p r in AB is 

r Me , a 

P '-.--.^ Bl -. 

Similarly, the maximum stress p rf in CD is 

*"-'* 

Consequently, / 



or, since from equation (86) ^ = - 



JU/2 

y 



By hypothesis, a > &. Therefore p" > y ; that is to say, the maxi- 
mum stress occurs in the strip CD, or in the direction of the shorter 
axis of the ellipse. In an elliptical plate, therefore, rupture may be 
expected to occur along a line parallel to the major axis, a result 
which has been confirmed by experiment. 

127. Maximum stress in homogeneous elliptical plate under 
uniform load. The method of finding the maximum stress in an 
elliptical plate is to consider the two limiting forms of an ellipse, 
namely, a circle and a strip of infinite length, and express a continu- 
ous relation between the stresses for these two limiting forms. The 
method is therefore similar to that used in Article 88 in obtaining 
the modified form of Euler's column formula. 

Consider first an indefinitely long strip with parallel sides, sup- 
ported at the edges and bearing a uniform load of amount w per unit 
of area. Let the width of the strip be denoted by 2 b, and its thickness 



184 STRENGTH OF MATERIALS 

by h. Then, if this strip is cut into cross strips of unit width, each 
of these cross strips can be regarded as an independent beam, the 
load on one of these unit cross strips being 2 ~bw, and the maximum 

(2 5) 2 w 

moment at the center being - '- -- Consequently, the maximum 

8 

stress in the cross strips, and therefore in the original strip, is 

4&V h 

,-s-V"fr 

12 ' 

In the preceding article it was shown that the maximum stress in 
an elliptical plate occurs in the direction of the minor axis. There- 
fore equation (87) gives the limiting value which the stress in an 
elliptical plate approaches as the ellipse becomes more and more 
elongated. 

For a circular plate of radius b and thickness h the maximum 
stress was found to be 

(S8) ,- 

Comparing equations (87) and (88), it is evident that the maximum 
stress in an elliptical plate is given, in general, by the formula 



where & is a constant which lies between 1 and 3. Thus, for - = 1, 

~b a 

that is, for a circle, k = 1 ; whereas, if - 0, that is, for an infinitely 

CL 

long ellipse, k = 3. The constant k may therefore be assumed to 
have the value 

&=3-2-> 
a 

which reduces to the values 1 and 3 for the limiting cases, and in 
other cases has an intermediate value depending on the form of the 
plate. Consequently, 



FLAT PLATES 



185 



which is the required formula for the maximum stress p in a homo- 
geneous elliptical plate of thickness h and semi-axes a and &. 

Problem 252. A cast-iron manhole cover 1 in. thick is elliptical in form, and 
covers an elliptical opening 3 ft. long and 18 in. wide. How great a uniform pres- 
sure will it stand ? 

128. Maximum stress in homogeneous square plate under uniform 
load. In investigating the strength of square plates the method of 
taking a section through the center of the plate and regarding the 
portion of the plate on one side of this section as a cantilever is used, 
but experiment is relied upon to determine the position of the dan- 
gerous section. From numerous experiments on flat plates, Bach 
has found that homogeneous square plates under uniform load always 
break along a diagonal.* 

Consider a homogeneous square plate of thickness h and side 
2 a, which bears a uniform load w per unit of area. Suppose that 
a diagonal section of this plate is taken, 
and consider either half as a cantilever, 
as shown in Fig. 120. Then the total load 
on the plate is 4wa 2 , and the reaction of 
the support under each edge is wa\ If d 
denotes the length of the diagonal AC, the 
resultant pressure on each edge of the plate 

is applied at a distance from AC, and 
therefore the moment of these resultants 




FIG. 120 



about A C is 2 (wa 2 ) - > or 



The total load on the triangle ABC 



is 2 wa 2 , and its resultant is applied at the center of gravity of the 
triangle, which is at a distance of - from AC. Therefore the mo- 

7 2>-7 

ment of the load about AC is (2 wa 2 ) - > or ^ - Therefore the total 

6 3 

external moment M at the section A C is 

wa 2 d wa 2 d wa 2 d 



* Bach, Elasticitat u. Festigkeitslehre, 3d ed., p. 561. 



186 



STRENGTH OF MATERIALS 



Hence the maximum stress in the plate is 

h 
2 



from which 
(90) 




D 




The maximum stress in a square plate of side 2 a is therefore the 
same as in a circular plate of diameter 2 a. 

Problem 253. What must be the thickness of a wrought-iron plate covering an 
opening 4 ft. square to carry a load of 200 lb./ft. 2 with a factor of safety of 5 ? 

129. Maximum stress in homogeneous rectangular plate under 
uniform load. In the case of rectangular plates experiment does 
Sa not indicate so clearly the posi- 

tion of the dangerous section as it 
does for square plates. It will be 
assumed in what follows, however, 
that the maximum stress occurs 
along a diagonal of the rectangle. 
This assumption is at least ap- 
proximately correct if the length 
of the rectangle does not exceed 
two or three times its breadth. 

Let the sides of the rectangle 
be denoted by 2 a and 2 b, and 
the thickness of the plate by h 
(Fig. 121). Also let d denote the 
length of the diagonal AC, and c 
the altitude of the triangle ABC. Now suppose that a diagonal section 
AC of the plate is taken, and consider the half plate AB C as a canti- 
lever, as shown in Fig. 121. If w denotes the unit load, the total 
load on the plate is 4 abw, and consequently the resultant of the 
reactions of the supports along AB and BC is of amount 2 abw, and 

/> 

is applied at a distance from A C. Therefore the moment of the sup- 
porting force about A C is abwc. Also, the total load on the triangle 




FIG. 121 



FLAT PLATES 187 

ABC is 2 abw, and it is applied at the center of gravity of the triangle, 

which is at a distance of - from A C. Consequently, the total moment 
o 

of the load about A C is - -- Therefore the total external moment 

M at the section A C is 

2 abwc abwc 
M abwc - = - > 
o o 

and the maximum stress in the plate is 

abwc li 
_Me__ 3 ' 2 _ 2 wale 

p '' i '' dtf ~dhT 

12 

or, since cd = 4 ab, 



which gives the required maximum stress. 

For a square plate a = b and c = a V2, and formula (91) reduces 
to formula (90) for square plates, obtained in the preceding article. 

Problem 254. A wrought-iron trapdoor is 5 ft. long, 3 ft. wide, and f in. thick. 
How great a uniform load will it bear ? 

130. Non-homogeneous plates ; concrete-steel floor panels. The 

formulas derived in the preceding articles apply only to flat plates of 
homogeneous material. If a plate is composed of non-homogeneous 
material, such as reenforced concrete, the maximum stress is given by 
the formula - , 



where /' is the moment of inertia of the equivalent homogeneous 
section obtained from the non-homogeneous section as explained in 
Article 48, and e' is the distance of the extreme fiber of this equivalent 
homogeneous section from its neutral axis. 

Thus, from Article 124, the external moment M on half of a uni- 

*7* ijf) 

f ormly loaded circular plate is M = > and, consequently, the maxi- 

o 

mum stress in a uniformly loaded, non-homogeneous, circular plate is 
given by the formula 



188 STRENGTH OF MATERIALS 

where I 1 and e f refer to the equivalent homogeneous section as 
explained above, and this section is taken through the center of 
the plate. 

Similarly, from Article 128, the maximum stress in a uniformly 
loaded, non-homogeneous, square plate of side 2 a is given by the 
formula 

/no\ 

(93) , 



and, from Article 129, the maximum stress in a uniformly loaded, non- 
homogeneous, rectangular plate of sides 2 a and 2 I by the formula 

abwce' 



(94) p = 



31' 



in which e r and /' refer to the equivalent homogeneous section obtained 
from a diagonal section of the plate. 

Problem 255. A concrete-steel floor panel is 18 ft. long, 15 ft. wide, and 4 in. 
thick, and is reenforced by square wrought-iron rods 1 in. thick, placed of an 
inch from the bottom of the slab and spaced 1 ft. apart. Find the maximum stress 
in the panel under a total live and dead load of 150 lb./ft. 2 . 

NOTE. Take a diagonal section of the panel and calculate the equivalent homogeneous 
section corresponding to it. Then find the position of the neutral axis of this equivalent 
homogeneous section, and its moment of inertia about this neutral axis, as explained in 
Article 48. The maximum stress can then be obtained from formula (94). 

Problem 256. Design a floor panel 14 ft. square, to be made of reenforced 
concrete and to sustain a total uniform load of 120 lb./ft. a with a factor of safety 
of 4. 



EXERCISES ON CHAPTER VHI 

Problem 257. The steel diaphragm separating two expansion chambers of a 
steam turbine is subjected to a pressure of 150 lb./in. 2 on one side and 801b./in. 2 
on the other. Find the required thickness for a factor of safety of 10. 

Problem 258. The cylinder of an hydraulic press is made of cast steel, 10 in. 
inside diameter, with a flat end of the same thickness as the walls of the cylinder. 
Find the required thickness for a factor of safety of 20. Also find how much larger 
the factor of safety would be if the end was made hemispherical instead of flat. 

Problem 259. The cylinder of a steam engine is 16 in. inside diameter and 
carries a steam pressure of 125 lb./in. 2 If the cylinder head is mild steel, find its 
thickness for a factor of safety of 10. 

Problem 260. A cast-iron valve gate 10 in. in diameter is under a pressure 
head of 200 ft. Find its thickness for a factor of safety of 15. 



FLAT PLATES 189 

Problem 261. A cast-iron elliptical manhole cover is 18 in. x 24 in. in size and is 
designed to carry a concentrated load of 1000 Ib. If the cover is ribbed, how thick 
must it be for a factor of safety of 20, assuming that the ribs double its strength ? 

Problem 262. Thurston's rule for the thickness of cylinder heads for steam 
engines is 

h = .00035 wD, 

where h = thickness of head in inches, 

1} = inside diameter of cylinder in inches, 
w = pressure in lb./in. 2 

Compare this formula with Bach's, assuming the material to be wrought iron, and 
using the data of Problem 259. 

Problem 263. Show that Thurston's rule for thickness of cylinder head, given 
in Problem 262, makes thickness of head =11 times thickness of walls. 

Problem 264. Nichols's rule for the proper thickness of unbraced flat wrought- 
iron boiler heads is 

_ Fw 

"Top' 

where h = thickness of head in inches, 

jP area of head in square inches, 
w = pressure per square inch, 

44,800 ult. strength in tension 



p = working stress = 



8 factor of safety 



Compare this empirical rule with Bach's formula, using the data of Problem 259 
and assuming the material to be wrought iron. 

Problem 265. Nichols's rule for the collapsing pressure of unbraced flat wrought- 
iron boiler heads is 

_ 10 hut 

where w = collapsing pressure in lb./in. 2 , 
h = thickness of head in inches, 
u t = ultimate tensile strength in lb./in. 2 , 
F area of head in square inches. 

Show that Nichols's two formulas are identical and that therefore they cannot be 
rational. 

Problem 266. The following data are taken from Nichols's experiments on flat 
wrought-iron circular plates. 

DIAMETER 

IN. 

34.6 
34.5 

28.5 
26.5 

Using this data, compare Bach's and Grashof's rational formulas with Nichols's 
and Thurston's empirical formulas, as given below: 



THICKNESS 


ACTUAL BURSTING 


IN. 


PRESSURE LB./IN.* 


A 


280 


I 


200 


1 


300 


1 


370 



1 ( JO STRENGTH OF MATEKIALS 

Circular plate, supported at edge and uniformly loaded. 



Bach, 

P 



Grashof , h== .4564 , 
\ Qp \p 

Nichols, ^=^ 
lOp 

Thurston, A = .00035 w>Z>, 
where /i = thickness of head in inches, 

D = diameter of head in inches = 2 r, 

w = pressure in lb./in. 2 , 

p = working stress in lb./in. 2 , 

F = area of head in square inches = 
Note that the Nichols and Thurston formulas apply only to wrought iron. 



CHAPTER IX 



CURVED PIECES: HOOKS, LINKS, AND SPRINGS 

131. Erroneous analysis of hooks and links. In calculating the 
strength of a curved piece whose axis is a plane curve, such as a hook 
or a link of a chain, many engineers are accustomed to assume that 
the distribution of stress is the same as in a straight beam subjected 
to an equal bending moment and axial load. For example, in calcu- 
lating the strength of a hook, such as shown in Fig. 122, the practice 
has been to take a section AB where the 

bending moment is a maximum, and cal- 
culate the unit stress p on AB by the 
formula p (pd)g 

P = h 

F I 

where the first term denotes the direct 
stress on the section AB of area F, and 
the second term represents the bending 
stress due to a moment Pd calculated 
from the formula for straight beams. 

The bending formula for straight 
beams, however, does not apply to curved 
pieces, as will be shown in what follows. 
Moreover, experiment has conclusively shown that a curved piece 
breaks at the point of sharpest curvature, whereas the above formula 
takes no account whatever of the curvature. The above formula is 
therefore not even approximately correct, and is cited as a popular 
error against which the student is warned. 

132. Bending strain in curved piece. Consider a curved piece 
which is subjected to pure bending strain, and assume that the axis 
of the piece is a plane curve and also that the radius of curvature is 
not very large as compared with the thickness of the piece. Hooke's 
law and Bernoulli's assumption will be taken as the starting point 

191 




FIG. 122 



192 



STRENGTH OF MATERIALS 



for the analysis of stress, as in the theory of straight beams ; that is 
to say, it will be assumed that the stress is proportional to the 
deformation produced, and that any plane section remains identical 
with itself during the deformation. 

Since the fibers on the convex side are longer than those on the 
concave side, it will take less stress to deform them an equal amount. 
Therefore the neutral axis does not pass through the center of gravity 
G of the section, but through some other point D, below G, as shown 
in Fig. 123. For if the neutral axis passed through G, the total 

deformation above and below 
G would be of equal amount, 
and therefore the total stress 
above G would be less than 
that below G, since the fibers 
above G are longer than those 
below. This shifting of the 
neutral axis constitutes the 
fundamental difference be- 
tween the theory of straight 
and curved pieces. 

Now let the length of any 
fiber, such as MN in Fig. 123, 

be denoted by Z, and the distance of this fiber from a gravity axis GZ 
by y. Also, let p denote the radius of curvature OG of the piece, j3 
the angle between two plane sections, and a the angle of deformation 
of a plane section. Then 

1 = /3>MO = (OG + GN)/3=(p + y)/3 t 
and the deformation dl of the fiber MN is 

dl = NN' = a ND = (y + d) a, 

where d denotes the distance GD between the neutral axis and the 
gravity axis. From Hooke's law, 




v Jit 

_ Edl _ E(y + d)a 

~~r (o- 



whence 



Let = Jc, where & is a constant. Then this expression for p reduces to 
P 



HOOKS, LINKS, AND SPRINGS 193 

(95) p^Ek^t+A. 

y + p 

Under the assumption of pure bending strain the shear is zero 
and the normal stresses form a couple. Therefore the algebraic sum 
of the normal stresses is zero ; that is to say, 



pdF = 0, 
or, substituting the value of p from equation (95), 



y + p 

Since k and E are constants and not zero, the integral must be zero. 
Therefore, separating the integral into parts, 



y + p 

whence 



r 

OB) d=- J y+p 



dF 

+ P 

which gives the distance of the neutral axis below the center of 
gravity of the section. 

Now let M denote the external bending moment acting on any 
given section of area F, dF an infinitesimal area taken anywhere in 
this section, p the stress acting on it, and y its distance from the 
gravity axis GZ. Then 



or, substituting the value of p from equation (95), 



y 

consequently 

k = 



y + p 

and hence 



194 



STRENGTH OF MATERIALS 



which is the required formula for calculating the bending stress at 
any point of a curved piece. 

133. Simplification of formula for unit stress. In formulas (96) 
and (97), derived in the preceding article, the integrals involved 
make the formulas difficult of application. The following geometrical 
transformation, which is due to Resal,* greatly simplifies the formulas 
and their application. 

The first step is a geometrical transformation of the boundary of 
the given cross section. Consider a symmetrical cross section, for 
example the circle shown in Fig. 124, and let OF be an axis of 

symmetry passing through the center 
of curvature C of the section, and OZ 
a gravity axis perpendicular to OY. 
Now suppose radii drawn from C to 
each point M in the boundary of the 
cross section. From H, the point of 
intersection of CM with the gravity 
axis OZ, erect a perpendicular to OZ, 
and from M draw a perpendicular 
to OY. Then these two perpendicu- 
lars will intersect in a point of the 
transformed boundary, as shown in 
Fig. 124. 

It will now be proved (1) that the 
distance of the center of gravity G 
of the transformed section from the 

center of gravity of the original section is the value of d given by 
formula (96), and (2) that the moment of inertia of the transformed 
section is the integral which occurs in formula (97). 

In Fig. 124 the distance NM' is the ^-coordinate of the point M ' 
let it be denoted by z f . Then 




NM' = z' = OH = MN = z 



CN p + y 
The distance d' of the center of gravity G of the transformed 



Resistance des Mate'riaux, pp. 385 et seq. 



HOOKS, LINKS, AND SPRINGS 195 

section below the center o'f gravity of the original section is 

zy - dy 



C z ' d y Cz-f 

J J p + y 



dy 



Dividing out the constant p and replacing the element of area zdy 
by dF, this expression for d' becomes 




y + p 

which is identical with the value of d given by formula (96) above. 
Consequently, the neutral axis of the original cross section coincides 
with the gravity axis of the transformed section. 

Now let the moment of inertia of the transformed section be 
denoted by I'. Then 



in which y' is measured from the gravity axis of the transformed 
section, that is, from a line through G parallel to OZ ; and dF' denotes 
an element of area of the transformed section ; whence dF ! = zdy'. 
Therefore, since 

y' = y + d, z' = z - > and dy' = dy, 

p + y 

the expression for /' becomes 



or, if the element of area zdy is denoted by dF, 

!< = p r(y + ^dF 
J p + y 

This integral, however, is the one which occurs in formula (97). 
Consequently, if its value from the above equation is substituted in 
(97), the expression for the unit stress p simplifies into 



196 



STEENGTH OF MATEEIALS 



For an ordinary beam without initial curvature, d = 0, I r = J, and 



p = oo, in which case, since 



= 1- 



y 



y+p p y 

My 



, the formula reduces to 



the ordinary beam formula p = - 

To avoid the confusion which may arise from positive and negative 
values of y in applying formula (98), note that 

y + d distance of fiber from neutral axis 

y + p distance of same fiber from center of curvature 

This quotient is then an abstract number, and its substitution in 

formula (98) gives the numerical value 
-I of the stress p without regard to sign. 

The following problem illustrates the 
application of the formula. 

Problem 267. The wrought-iron crane hook, 
shown in Fig. 125, is designed to support a load 
of ten tons. Find the maximum stress in the 
hook under this load, and thence determine the 
factor of safety. 

Solution. Let a cross section OCF of the hook 
be taken at the position of maximum moment, 
as shown in the shaded projection in Fig. 125. 

In Fig. 126 let the curve numbered 1 repre- 
sent this projection. The gravity axis DF of 
this section, perpendicular to the axis of sym- 
metry COF, is first determined, which may be 
done by the graphical method explained in 
Article 47, or otherwise.* Curve 1 is then trans- 
formed into curve 2 by the method explained in 
Article 133, the light construction lines on the 
left of OF showing how this is accomplished. 

The moment of inertia I' of curve 2 is then 

found graphically by the method explained in Article 47. This method consists 
in first transforming curve 2 into curves 3 and 4, as there explained, then measuring 
the areas between OF and curves 3 and 4 by means of a planimeter, and finally sub- 
stituting the areas so found in the formulas for the moment of inertia I' of curve 
2 and the distance c of its center of gravity from AB, given in Article 47. 

In the present case we have then the following numerical values for substitution : 
p=C# = 4.4in., F=7.9in. 2 , CO = 2.2 in., EO' = 2.8 in., 

M= 20,000 x 4.4 = 88,000 in. Ib. 

* A simple method of determining a gravity axis sufficiently accurate for ordinary pur- 
poses consists in cutting the section out of cardboard and balancing it on a knife edge. 




FIG. 125 



HOOKS, LINKS, AND SPRINGS 



197 



Consequently, at the outer fiber O' we have 

88,000 x 4.4 2.8 + 0.4 

x = 1 2,200 lb./ in. 2 compression; and at the inner fiber 0, 



P = 



p = 



14.1 

88,000 x 



_ 



= - 22,470 lb./i,,., tension. 



Moreover, the direct tensile stress on the cross section is 



20,000 
7.9 



= 2530 lb./in.' 



Hence the actual total stress on the outer fiber O' is 12,200 2530 = UG70 lb./in. s 
compression, corresponding to 
a factor of safety of about 5; 
and on the inner fiber is 
22,470 + 2530 = 25,000 lb. /in. 2 
tension, corresponding to a fac- 
tor of safety of 2. 

Problem 268. By the for- 
mula given in Article 131, cal- 
culate the maximum bending 
stress and the maximum total 
stress on the hook shown in 
Fig. 125, and compare the re- 
sults with those of the preced- 
ing problem. 

Problem 269. The danger- 
ous section of a hook similar to 
that shown in Fig. 125 has for its 
dimensions b = 2| in., h = Q in., 
ri = If in., r 2 = fin. (Fig. 127), 
and OC = 2| in. (Fig. 126). Using 
a factor of safety of 4, find the 
safe load for the hook. 

For all practical purposes the 
theory of stress in curved pieces 
here presented is undoubtedly 
the most satisfactory theory 
which has yet been developed. 
A more rigorous analysis of the 
subject, however, introducing 
Poisson's ratio of lateral def- 
ormation, has been given by 
Andrews and Pearson in their monograph on crane and coupling hooks.* Although 
this discussion is extremely valuable , from a theoretical standpoint, it has been 
shown that its results exhibit but a slight refinement over the simpler discussion 
given above, a difference considerably less than the variation which may be ex- 
pected in the physical properties of materials used commercially.! By reason of 

* Karl Pearson and E. H. Andrews, "Theory in Crane and Coupling Hooks," etc., 
Tech. Series I; J)rapers Co. Research Memoirs [Dulau and Co., 37 Soho Square, London, W.], 
See also Am. Much., Vol. XXXII, Oct. 7, 1909, pp. 615-619; Dec. 16, 1909, pp. 1065-1067. 

t Am. Much., Vol. XXXIII, Nov. 24, 1910, pp. 954-955. 




198 



STRENGTH OF MATERIALS 



this uncertainty as to the exact values of the physical constants involved, the 
simpler method is of more value to the designer. The enormous amount of labor 
and liability to error involved in the application of the Pearson- Andrews formula 
is, in fact, prohibitive where speed and accuracy are an object. 

* 134. Curved piece of rectangular cross section. If the cross 
section of a curved piece is rectangular, the integrals in formulas (96) 
and (97), Article 132, can be easily evaluated. These formulas may 
therefore be used for calculating the strength of the piece in prefer- 
ence to the graphical method explained in the preceding article. 

Let the cross section of the piece be a rec- 
tangle of breadth b and depth h, and let p 
denote the radius of curvature of the 1 piece 
at the section under consideration. From 
formula (96), the distance of the neutral axis 
of the section from the mean fiber, or gravity 
axis, is 

J I 





y + p 



where y denotes the distance of the infinitesi- 
mal area dF from the gravity axis. In the 
present case dF = Idy ; hence 



n n 

> C 2 ydy r* 

I , y -\- n I , 

is h u ' i is h 



FIG. 127 



f* dy r\ 

I 7 H -h D / i ' 

*/ h,y ' r i/ hi 



dy 



By* division, - = 1 Consequently, the numerator of the 

y+p y+p 

above fraction becomes 



r 

J 



* For a brief course the remainder of this chapter may be omitted. 



HOOKS, LINKS, AND SPRINGS 

Similarly, the denominator becomes 



19!) 



Consequently, 



8 dy 
y + p 



d=- 



+ p) 



2 p h 



"2p-h 



h 



which may be written 

(99) d = p - 



> e 2p-h 
h 



h 



2p-h 

From formula (97), Article 132, the unit stress^? at any point in the 
cross section, distant y from the mean fiber, is given by the equation 

M(y + d) 



(y + p) 



(y 



dF 



Replacing dF by ~bdy, and separating the integral in the denominator 
into partial integrals by means of division, this integral becomes 






- 



Substituting for d its value from equation (99), this expression 
finally reduces to 



y + p 



p- 



2/o hj 



= bhd. 



STRENGTH OF MATERIALS 



Hence the expression for p becomes 



(100) 



The stresses on the extreme fibers are the values of p for y = 
Hence 

(101) 




FIG. 128 



(7*, 2/9) bhd 

Note that the stress on the inside fiber is always negative, in con- 
sequence of which the sign of M should be 
negative if it tends to decrease the radius of 
curvature, and vice versa. 

Problem 270. A boat's davits are composed of 
two wrought-iron bars 2| in. square, bent to a radius 
of 2 ft., as shown in Fig. 128. If the boat weighs 
500 Ib. and is hung 3} ft. from the vertical axis of 
the davits, find the maximum stress in the davits 
and the factor of -safety. 

135. Effect of sharp curvature on bending 
strength. Consider a sharply curved pris- 
matic piece which is subjected to bending strain. From the above 
discussion, it is known that for a section taken in the neighborhood of 
the bend, the neutral axis does not coincide with the gravity axis but 
approaches the center of curvature. The neutral 
fiber is therefore separated from the mean fiber, 
or axis of the piece, and takes some such posi- 
tion as that shown by the broken line in Fig. 129. \ 
Consequently the inner fiber through A must \ \\ 
endure a far greater stress than that deduced 
from formulas for the straight portion. Engi- 
neers and constructors have learned by experi- 
ence that sharp curvature produces weakness 
of this kind, and that it is necessary to reenforce a piece at a bend 
either by increasing its diameter or by adding a brace. 

As an illustration of the effect of sharp curvature on bending 
strength, suppose that a bar of rectangular cross section is bent into a 
right angle, as shown in Fig. 130. In this case the center of curvature 



FIG. 129 



HOOKS, LINKS, AND SPRINGS 



201 



of the mean fiber BC is at A. Therefore, if h denotes the thickness 
of the piece, the radius of curvature of BC is 

p = - Consequently, 



2 + h 



2k 



and hence formula (99) becomes 

h 
d = p = -. 




FIG. 130 



Therefore the neutral fiber passes through the ver- 
tex of the angle A, and consequently a piece of 
this kind can offer no resistance to bending. In other words, if a 
piece is bent exactly at right angles on itself, the slightest bending 
strain must produce incipient rupture. 

This example is useful, then, in pointing out the danger of sharp 
curvature and showing how rapidly the strength decreases with the 
radius of curvature. 

136. Maximum moment in circular piece. Consider a prismatic 
piece with a circular axis, such as a ring or a section of pipe, 

and suppose that it is 
subjected to two equal 
and opposite forces 
P, either of tension 
or compression, act- 
ing along a diameter 
as shown in Fig. 131. 
Draw a second diam- 
eter MN at right an- 
gles to the direction 
in which the forces P 
act. Since these two 
diameters divide the 

figure into four symmetrical parts, it is only necessary to consider one 
of these parts, say the upper left-hand quadrant. The forces acting on 

any section of this quadrant consist of a single force and a moment. 

p 
On the base CD of the quadrant this single force is of amount > 

2 




FIG. 131 



202 STRENGTH OF MATERIALS 

and the unknown moment will be denoted by Jf . On any other sec- 
tion AB the bending moment M and single force P' are respectively 

M=M*+j{p -pcaa/3), 

(102) p 

P' = - cos A 

in which p is the radius of the mean fiber and /? is the angle which 
the plane of the section AB makes with the base CD. 

Now, no matter whether the section is flattened or elongated by 
the strain, from the symmetry of the figure the diametral sections 
MN and PP will always remain at right angles to one another. 
Therefore the total angular deformation A/3 for the quadrant under 
consideration must be zero ; that is to say, 



But, from Article 67, 
Consequently, 



MI 



= 0. 



El 
Inserting in this expression the value of M obtained above, 



or 



pd/3=0, 




7T 

Jo \ 

whence 

TT A 



which is the maximum negative moment. 

From formula (102), the maximum positive moment must occur 

when cos ft = 0, that is, when /3 = > or at top and bottom. Therefore 

^na X = ^ = =-318 Pp. 
7T 



HOOKS, LINKS, AND SPRINGS 



203 



The maximum moment, therefore, occurs at the points of application 
of the forces. From formula (102), the direct stress at these points 
is zero. 

Having determined the position and amount of the maximum 
bending- moment, the maximum bending stress can be calculated by 
the graphical method explained in Article 133, or, if the piece is rec- 
tangular in section, by formulas (99) and (100) or (101) in Article 134. 

Problem 271. A wrought-iron anchor ring is 6 in. in inside diameter and 2 in. in 
sectional diameter. With a factor of safety of 4, find by the graphical method of 
Article 133 the maximum pull which the ring can 
withstand. 

Problem 272. A cast-iron pipe 18 in. in in- 
ternal diameter and 1 in. thick is subjected to 
a pressure of 150 Ib. /linear foot at the highest 
point of the pipe. Find the maximum stress in 
the pipe. 

HINT. Use formula (101), Article 134. 

137. Plane spiral springs. Consider a 
plane spiral spring, such as the spring of 
a clock or watch. Let P denote the force 
tending to wind up the spring, and c the 

perpendicular distance of P from the spindle on which the spring is 
wound (Fig. 132). Also, let dx denote a small portion of the spring 
at any point A distant y from P. Then the moment at A is M = Py ; 
and hence, from Article 6 7, the angular deformation dp for the portion 
dx is given by the formula 

-jo _ Mdx _ Pydx 

" ~ ~&T ~ T?T ' 

Jiil M/l 

Therefore the total angular deformation of the spring is 




FIG. 132 



Since the average value of y is c, and the integral of dx is the length 
of the spring I, ~ l 

I ydx = cl, 
Jo 
and hence 



7 



204 STRENGTH OF MATERIALS 

The resilience W of the spring is, therefore, 



If the spring is of rectangular cross section, which is the usual 
form for plane spiral springs, the stress can be calculated by formulas 
(99) and (101), Article 134. 

The formula just obtained for the resilience of a spring is a special 
case of a more general formula. Thus consider a portion of a beam of 
length AB = I, and let M denote the average bending moment over the 
part considered, and (3 the change in slope in passing from A to B. Then 

the work done in bending the 
portion AB is W= \M(3, or, since 

Ml ,, . M 2 l 

p = , this becomes W = 
El 2 El 

\_ In the case of the spring con- 
sidered above, the mean value 
of the bending moment was 
M=Pc. 

Furthermore, if p denotes the 
greatest stress at the elastic 
limit and e the distance at which" it acts from the neutral axis, 

then M= > and consequently the resilience of the beam is 



r~Lt 




FIG. 133 



For the resilience of a piece under direct stress, see Article 22. 

Problem 273. A steel clock spring | in. wide and ^ in. thick is wound on a 
spindle T 3 ^ in. in diameter. With a factor of safety of 5, what is the maximum 
moment available for running the mechanism ? 

Suggestion. The dangerous section occurs at the spindle where the moment is 
greatest and the radius least. Therefore, in the present case, p = T V 5 in., h = g \ in., 
b = | in. Also, since the ultimate tensile strength of spring steel is about 240,000 

lb./in.2, p m&x = 24 ' 00 = 48,000 lb./in. 2 d can then be calculated by formula (99), 
5 

and M by formula (101). 



HOOKS, LINKS, AND SPKINGS 



205 



EXERCISES ON CHAPTER IX 



Problem 274. A flat spiral spring is in. broad, -^ in. thick, and 12 ft. long. 
What is the maximum torque it can exert on a central spindle if the stress is not 
to exceed 60,000 lb./in. 2 ? 

Problem 275. The links of a chain are made of f-in. round wrought iron, with 
semi-circular ends of radius 1 in. Straight portion of link 1 in. long. Find the 
maximum stresses in the link due to a pull 
of 1 ton on the chain. 

Problem 276. A ring is made from a 
round steel rod 1 in. in diameter. The 
inside diameter of the ring is 6 in. Find 
the maximum stress resulting from a pull 
on the ring of f ton. 

Problem 277. Calculate the maximum 
tensile and compressive stresses on the 
cross section of the hydraulic riveter 
shown in Fig. 62, page 79. 

Problem 278. In Fig. 133 a design is 
shown for the cross section of a punch 
press frame.* Substitute this cross sec- 
tion for that shown in Fig. 62, page 79, 
calculate the maximum stresses, and com- 
pare with the results of Problem 277. 

Problem 279. The following table gives 
the dimensions of crane hooks for the 

design shown in Fig. 134 for loads from 5 to 50 tons.t Compute the maximum 
stresses at the dangerous section in each case and determine the factor of safety. 




T 



FIG. 134 



TONS 


A 


B 


c 


D 


E 


F 


H 


K 


M 


N 


Q 


R 


s 


T 


5 ... 


1| 


31 


3 2 


31 


41 


4 


3| 


l 

2 


2| 


3 


11 


31 


14 


1 


10 ... 


2f 


41 


5 


5 


6 


51 


5 


1 


31 


4 


21 


5 


11 


I 


15 ... 


3 


6* 


8* 


6 


7i 


4 


5f 


8 

4 


44 


4f 


2| 


6 


If 


5 

3 


20 ... 


8$ 


6 


6 1 , 


6| 


84 


7 


61 


! 


4| 


64 


34 


6| 


2 


1 


25 ... 


3 7 


6| 


71 


71 


94 


8 


7f 


1 


6i 


6 


8* 


7| 


24 


I 


30 ... 


4.1 


n 


8* 


84 


i4 


81 


84 


i 


6 


6| 


4 


81 


21 


! 


40 ... 


4f 


8| 


94 


i 


ii| 


10 


84 


i 


6| 


7| 


4i 


0| 


3 


i 


50 ... 


54 


<4 


11 


10J 


12 


114 


104 


i 


71 


84 


5 


101 


81 


14 



* Rautenstrauch, Am, Mach., December 16, 1909. 
t Dixon, Am. Mach., August 16, 1900. 




206 STRENGTH OF MATERIALS 

Problem 280. In Fig. 135 a design for the cross section of a crane hook is shown 
in which all the dimensions are expressed in terms of a single quantity r. The 

values of this constant r for various loads 
"~^ are given by the designer as follows :* 

, ^ , , "I 40-ton hook, r=o.54; 

30-ton hook, r = 4.7 ; 
20-ton hook, r = 3.94 ; 
10-ton hook, r = 2.76; 

5-ton hook, r = 1.95 ; 

2-ton hook, r = 1.23. 

Compare the strength of this design for a given load with that of the design shown 
in Fig. 134. 

NOTE. Materials like cast iron, which do not conform to Hooke's law, cannot be 
subjected to a rigorous stress analysis. For example, the Pearson-Andrews formula is 
based on Poisson's ratio, which is one of the most refined elastic properties, and it is 
therefore useless to attempt to calculate the stress in a casting by such formulas. More- 
over, it has recently been shown by experiment that the initial stresses due to cooling 
in an irregular casting, such as a punch or riveter frame, are so great as to upset any 
exact calculations of the bending stresses involved. t In these experiments many of the 
specimens failed by a vertical crack appearing in the web just back of the inner, or com- 
pression, flange, i.e. perpendicular to the section AB in Fig. 62, page 79, a form of failure 
which has no apparent relation to the theory of flexure. These experiments were also 
valuable in showing the practical necessity of putting a fillet in the corners where the 
web joins the inner flange, or increasing the thickness of the web at this point, as shown 
in Fig. 133. 

In many machine tools the rigidity of the frame is the factor which determines the 
design, rather than the strength of the construction. In all such cases empirical meth- 
ods based on practical experience are the ones that should be employed. 

* Rautenstrauch, Am. Mach., December 16, 1909. 

t A. L. Jenkins, "The Strength of Punch and Riveter Frames made of Cast Iron," 
Jour. Am. Soc. Mech. Eng., Vol. XXXII, pp. 311-332. 



CHAPTER X 



ARCHES AND ARCHED RIBS 

I. GRAPHICAL ANALYSIS OF FORCES 

138. Composition of forces. In determining the effect which a 
given system of forces has upon a body, it is often convenient to 
represent the forces by directed lines and calculate the result graphic- 
ally. In this method of representation the length of the line denotes 
the magnitude of the force laid off to any given scale, and the direc- 
tion of the line indicates the direction in which the force acts, or its 
line of action. 

When the lines of action of a system of forces all pass through 
the same point, the forces are said to be concurrent. The simplest 
method of dealing with such a system is to find the amount and line 
of action of a single force which would have the same effect as the 
given system of forces upon the motion of the point at which they 
act. This single force is called the resultant of the given system 
and its equal and opposite the equilibrant. When each of a system of 
forces acting on a body balances the others so that 
the body shows no tendency to move, the forces 
are said to be in equilibrium, in which case their 
resultant must be zero. 

The resultant of two forces acting at a point 
is found by drawing the forces to scale in both 
magnitude and direction, and constructing a 
parallelogram upon these two lines as adjacent 
sides ; the diagonal of this parallelogram is then the required resultant 
(Fig. 136). This construction can be verified experimentally by fas- 
tening a string at two points A and B and suspending a weight R 
from it at any point C (Fig. 137). Then if two forces equal in magni- 
tude to the tension in AC and BC are laid off parallel to AC and BC 

207 





FIG. 136 



208 



STRENGTH OF MATERIALS 




FIG. 137 




respectively, it will be found that their resultant is equal and par- 
allel to R, and opposite in direction. 

Since the opposite sides of a parallelogram 
are equal and parallel, it is more convenient in 
finding the resultant of two forces to construct 
half the parallelogram. Thus, in the preceding 
example, if P 2 is laid off from the end of P v 
R is the closing side of the triangle so formed 
(Fig. 138). Such a figure is called a force triangle. 
In order to find the resultant of several con- 
current forces lying in the same plane, it is 
only necessary to comhine two of 
them into a single resultant, com- 

bine this resultant with a third force, and so on, taking 
the forces in order around the point in which they meet. 
Thus, in Fig. 139, E l is the resultant of P l and P 2 ; R z is 
the resultant of R l and P s ; R s is the resultant of R z 
and P 4 ; and R is the resultant of R z and P 5 . R is there- 
fore the resultant of the entire system P I} P 2 , P 3 , P 4 , P 5 . 

In carrying out this construction it is unnecessary to draw the 

intermediate resultants 
RV R 2 , and P 3 , the final 
resultant in any case 
being the closing side of 
the polygon formed by 
placing the forces end 
to end in order. Such 
a figure is called a force 
polygon. From the above 
construction it is evi- 
dent that the necessary 
and sufficient condition 
that a system of concur- 
rent forces shall be in 
equilibrium is that their 
force polygon shall close, since in this case their resultant must 
be zero. 



FIG. 138 




FlG ' 139 



ARCHES AND ARCHED RIBS 



209 



The resultant of a system of non-concurrent forces lying in the 
same plane, that is to say, forces whose lines of action do not all pass 
through the same point, is found by means of a force polygon as 
explained above. In this case, however, the closing of the force 
polygon is not a sufficient condition for equilibrium, for the given 
system may reduce to a pair of equal and opposite forces acting in 
parallel directions, called a couple, which would tend to produce rota- 
tion of the body on which they act. For non-concurrent forces, there- 
fore, the necessary and sufficient conditions for equilibrium are first, 
the resultant of the given system must be zero, and second, the sum of 
the moments of the forces about 
any point must be zero. 

Suppose that the force polygon 
corresponding to any given system 
of forces is projected upon two 
perpendicular lines, say a vertical 
and a horizontal line. Then since 
the sum of the projections upon 
any line of all the sides but one 



D 

F " 

c"lc 




1) 




C'B'D' 



A'E' 
FIG. 140 



F'G' 



of a polygon is equal to the pro- 
jection of this closing side upon 
the given line, the sum of the horizontal projections of any system 
of forces is equal to the horizontal projection of their resultant, and 
the sum of their vertical projections is equal to the vertical projection 
of their resultant (Fig. 140). 

The conditions for equilibrium of a system of forces lying in the 
same plane may then be reduced to the following convenient form. 

1. For equilibrium against translation, 

I ^y horizontal components = O, 
I V vertical components = O. 

2. For equilibrium against rotation, 

Vmoments about any point = O. 

If the forces are concurrent, rotation cannot occur, and the first 
condition alone is sufficient to assure equilibrium. In order that 



210 



STRENGTH OF MATERIALS 



a system of non-concurrent forces shall be in equilibrium, how- 
ever, both conditions must be 
fulfilled. 



6 lb. 




Problem 281. Construct the re- 
sultant of the system of concurrent 
forces shown in Fig. 141. 

Problem 282. Determine 
whether or not the system of par- 
allel forces shown in Fig. 142 satis- 
fies conditions 1 and 2 above. 

139. Equilibrium polygon. 

The preceding construction for 
the force polygon gives a method for calculating the magnitude and 



FIG. 141 



3 tons 



- 4-' 



2 tons 



3^ tons 



2 1 tons 



4 tons 
FIG. 142 



direction of the result- 
ant of any given system 
of forces, but does not 
determine the line of 
action of their resultant. 
The most convenient 3-tons 
way to determine the 
line of action of the re- 
sultant is to introduce into the given system two equal and opposite 

1 forces of arbi- 
trary amount 
and direction, such 
as P' and P" in 
Fig. 143 (A). 

Since P f and P" 
balance one an- 
other, they will not 
affect the equilib- 
rium of the given 
system. This is 
obvious from the 
force polygon. For 
in Fig. 143 (B), let 
R denote the resultant of the given system of forces P l P 4 . Then, 
if A represents in magnitude and direction the arbitrary force P', OB 




ARCHES AND ARCHED RIBS 



211 



is the resultant of P' and P 1? OCis the resultant of OB and P 2 , etc., and 
finally OE, or P'", represents the resultant of P 1 ', P x , P 2 , P 8 , P 4 . If 
then P'" is combined with P", the resultant JK is obtained as before. 

Now to find the line of action of R, suppose that P' and P l are 
combined into a resultant R^ acting in the direction B'A' (Fig. 143 (C)) 
parallel to the ray OB of the force polygon (Fig. 143 (B)). Prolong 
A'B' until it intersects P 2 , and then combine R^ and P 2 into a result- 
ant R z acting in the direction C'B' parallel to the ray OC of the force 
polygon. Continue in this manner until P'" is obtained. Then the 
resultant of P" and P' n will give both the magnitude and line of 




FIG. 144 



action of the resultant of the original system P 1? P 2 , P 8 , P 4 . The 
closed figure A'B' C'D'E'F' obtained in this way is called an equilibrium 
polygon. 

For a system of parallel forces the equilibrium polygon is con- 
structed in the same manner as above, the only difference being that 
in this case the force polygon becomes a straight line, as shown in 
Fig. 144. 

Since P' and P" are entirely arbitrary both in magnitude and 
direction, the point 0, called the pole, may be chosen anywhere in the 
plane. Therefore, in constructing an equilibrium polygon correspond- 
ing to any given system of forces, the force polygon ABODE (Fig. 143) 
is first drawn, then any convenient point is chosen and joined to 
the vertices A, B, C, D, E of the force polygon, and finally the equi- 
librium polygon is constructed by drawing its sides parallel to the 
rays OA } OB, OC, etc., of the force diagram. 



212 



STEENGTH OF MATEEIALS 



Since the position of the pole is entirely arbitrary, there is an 
infinite number of equilibrium polygons corresponding to any given 
set of forces. The position and magnitude of the resultant R, how- 
ever, is independent of the choice of the pole, and will be the same, 
no matter where is placed. 

Problem 283. The ends of a cord are fastened to supports and weights attached 
at different points of its length. Show that the position assumed by the string is 
the equilibrium polygon for the given system of loads. 

140. Application of equilibrium polygon to determining reactions. 

One of the principal applications of the equilibrium polygon is in 
determining the unknown reactions of a beam or truss. To illustrate 
its use for this purpose, consider a simple beam placed horizon- 
tally and bearing a number of vertical loads P lt P 2 , etc. (Fig. 145). 
To determine the reactions E^ and R z , the force diagram is first 





FIG. 145 



constructed by laying off the loads P lt P 2 , etc., to scale on a line AF, 
choosing any convenient point as pole and drawing the rays OA, 
OB, etc. The equilibrium polygon corresponding to this force diagram 
is then constructed, starting from any point, say A 1 , in R r 

Now the closing side A'G' of the equilibrium polygon determines 
the line of action of the resultants P' and P" at A r and G r respectively. 
For a simple beam, however, the reactions are vertical. Therefore, in 
order to find these reactions each of the forces P' and P' f must be 
resolved into two components, one of which shall be vertical. To 
accomplish this, suppose that a line OH is drawn from the pole in 



AKCHES AND ARCHED EIBS 



213 



the force diagram parallel to the closing side G r A' of the equilibrium 
polygon. Then HO (or P') may be replaced by its components HA 
and AO, parallel to E 1 and A'B 1 respectively ; and similarly, OH may 
be replaced by its components FH and OF, parallel to jft 2 and F' G' 
respectively. HA and FH are therefore the required reactions. 

Problem 284. A simple beam 20 ft. long supports concentrated loads of 3, 5, 2, 
and 9 tons at distances of 5, 7, 14, and 18 ft. respectively from the left support. 
Calculate the reactions of the supports graphically. 

Problem 285. Construct an equilibrium polygon for a simple beam bearing a 
uniform load, and show that the reactions are equal. 

141. Equilibrium polygon through two given points. Let it be 

required to pass an equilibrium polygon through two given points, 
say M and N (Fig. 146). 

To solve this problem a trial force diagram is first drawn with any 
arbitrary point as pole, and the corresponding equilibrium polygon 




FIG. 146 

MA'B'C'D'E' constructed, starting from one of the given points, say 
M. The reactions are then determined by drawing a line OH parallel 
to the closing side ME' of the equilibrium polygon, as explained in 
the preceding article. 

The reactions, however, are independent of the choice of the pole 
in the force diagram, and consequently they must be of amount AH 
and HE, no matter where is placed. Moreover, if the equilibrium 
polygon is to pass through both M and N, its closing side must coin- 
cide with the line MN t and therefore the pole of the force diagram 
must* He somewhere on a line through H parallel to MN. Let O 1 be 



214 



STRENGTH OF MATERIALS 



a point on this line. Then if a new force diagram is drawn with 0' 
as pole, the corresponding equilibrium polygon starting at M will pass 
through N. 

142. Equilibrium polygon through three given points. Let it be 
required to pass an equilibrium polygon through three given points, 
say M, N, and L (Fig. 147). 

As in the preceding article, a trial force diagram is first drawn 
with any point as pole, and the corresponding equilibrium polygon 
constructed, thus determining the reactions R 1 and R 2 as AH and 
HE respectively. 

Now if the equilibrium polygon is to pass through N, the pole of 
the force diagram must lie somewhere on a line HK drawn through 




K 




1) 



FIG. 147 

H parallel to MN, as explained in the preceding article. The next 
step, therefore, is to determine the position of the pole on this line 
HK, so that the equilibrium polygon through M and N shall also pass 
through L. This is done by drawing a vertical LS through L and 
treating the points M and L exactly as M and N were treated. Thus 
GAB CD is the force diagram for this portion of the original figure, and 
MA'B'C'S is the corresponding equilibrium polygon, the reactions 
for this partial figure being H'A and DH'. If, then, the equilibrium 
polygon is to pass through L, its closing side must be the line 
ML, and consequently the pole of the force diagram must lie on a 
line H'K' drawn through H' parallel to ML. The pole is therefore 
completely determined as the intersection 0' of the lines HK and 
H'K' . If, then, a new force diagram is drawn with 0' as pole, the 



AKCHES AND AKCHED KIBS 



corresponding equilibrium polygon starting from the point M will 
pass through both the points L and N. 

Since there is only one position of the pole 0', but one equilibrium 
polygon can be drawn through three given points. In other words, an 
equilibrium polygon is completely determined by three conditions. 

143. Application of equilibrium polygon to calculation of stresses. 
Consider any structure, such as an arch or arched rib, supporting a 
system of vertical loads, and suppose that the force diagram and 
equilibrium polygon are drawn as shown in Fig. 148. Then each 
ray of the force diagram is the resultant of all the forces which pre- 
cede it, and acts along the segment of the equilibrium polygon parallel 
to this ray. For instance, OC is the resultant of all the forces on the 




FIG. 148 




D 



left of P 3 , and acts along C'D'. Consequently the stresses acting on 
any section of the structure, say mn, are the same as would result 
from a single force OC acting along C'D'. 

Let 6 denote the angle between the segment C'D' of the equilibrium 
polygon and the tangent to the arch at the point S. Then the stresses 
acting on the section mn at S are due to a tangential thrust of amount 
OC cos ; a shear at right angles to this, of amount OC sin ; and a 
moment of amount OC-d, where d is the perpendicular distance of 
C'D' from S. 

From Fig. 148, it is evident that the horizontal component of any 
ray of the force diagram is equal to the pole distance OH. There- 
fore if 0(7 is resolved into its vertical and horizontal components, the 
moment of the vertical component about S is zero, since it passes 



216 



STRENGTH OF MATERIALS 



through this point ; and hence the moment OC d = OH- z, where z is 
the vertical intercept from the equilibrium polygon to the center of 
moments S. Having determined the moment at any given point, the 
stresses at this point can be calculated as explained in Article 157. 

144. Relation of equilibrium polygon to bending moment diagram. 
In the preceding article it was proved that the moment acting at any 
point of a structure is equal to the pole distance of the force diagram 
multiplied by the vertical intercept on the equilibrium polygon from 
the center of moments. For a system of vertical loads, however, the 
pole distance is a constant. Consequently the moment acting on any 
section is proportional to the vertical intercept on the equilibrium 
polygon from the center of moments. Therefore, if the equilibrium 
polygon is drawn to such a scale as to make this factor of propor- 
tionality equal to unity, the equilibrium polygon will be identical 
with the bending moment diagram for the given system of loads. 

Problem 286. Compare the bending moment diagrams and equilibrium polygons 
for the various cases of loading illustrated in Article 62. 



II. CONCRETE AND MASONRY ARCHES 

145. Definitions and construction of arches. The following dis- 
cussion of the arch applies only to that form known as the barrel 
arch. Domed and cloistered arches demand a special treatment which 

is beyond the scope of 
this volume. 

The various portions of 
a simple, or barrel, arch, 
such as shown in projec- 
tion in Fig. 149, have the 
following special names. 

Soffit : the inner or con- 
cave surface of the arch. 

Intrados : the curve of in- 
tersection (ACB, Fig. 149) of the soffit, with a vertical plane perpendicular 
to the axis, or length, of the arch. 

Extrados : the curve of intersection (DEF, Fig. 149) of a vertical plane 
with the outer surface of the arch. 
Crown : the hk;-hc 't wrt of the arch. 




ARCHES AND AECHED BIBS 217 

Haunches : the parts of the arch next to the abutments. 

Springing line : the line AB joining the ends of the intrados. 

Rise : the distance from the springing line to the highest point of the 
intrados. 

Spandrel : the space above the extrados. In the case of an arch supporting 
a roadway, the filling deposited in this space is called the spandrel Jilting. 

Voussoir : any one of the successive stones in the arch ring of a masonry arch. 

Keystone : the central voussoir. 

In constructing an arch the material is supported while being put 
in place by a wooden structure called a center, the outer surface of 
which has the exact form of the soffit of the required arch. The 
center is constructed by making a number of frames or ribs having 
the form of the intrados of the required arch, and then placing these 
ribs at equal intervals along the axis of the arch and covering them 
with narrow wooden planks, called lagging, running parallel to the 
axis of the arch. When the arch is completed, or, in case of a con- 
crete arch, when the material has hardened sufficiently to resist the 
stress due to its weight, the centers are removed, thus leaving the 
arch self-supporting. 

146. Load line. Since the filling above an arch has the same form 
as the arch itself, it must be partly self-supporting. In designing an 
arch, however, no advantage is taken of this fact, and it is assumed 
that any portion of the extrados supports the entire weight of the 
material vertically above it. The only exception to this is in the 
construction of tunnel walls, in which case it would be obviously 
unnecessary as well as impracticable to construct an arch sufficiently 
strong to support the entire weight of the material above it. 

If the filling above an arch is not of the same material as the arch 
ring, subsequent calculations are greatly simplified by constructing a 
load line which shall represent at any point the height which a filling 
of the same material as the arch itself must have in order to produce 
the same load as that actually resting on the arch. The vertical 
intercept between the intrados and the load line will then represent 
the load at any given point of the arch. 

In case of a live load the load line will have a different form for 
each position of the moving load. 

Problem 287. A circular arch of 20 ft. span and 6 ft. rise, with an arch ring 
3 ft. thick, is composed of concrete weighing 140 lb./ft. 3 Construct the load line 



218 



STRENGTH OF MATERIALS 




for a roadway three feet above the crown of the arch, with a spandrel filling of 

earth weighing 100 lb.'/ft. 3 

Solution. In this case the weight of a cubic foot of the spandrel filling is to the 

weight of a cubic foot of the arch ring as 100 : 140. Therefore the load line is 

obtained by reducing the 
intercept on each ordinate 
between the roadway and 
the extrados in the ratio 
140:100. Thus, in Fig. 150, 
reducing any ordinate AB 
in this ratio we obtain the 
ordinate EC, etc. By car- 
rying out this reduction 
on a sufficient number of 
ordinates, and joining the 
points C so found, the load 
line DJECFG is obtained. 

147. Linear arch. Suppose that the voussoirs of an arch have 
slightly curved surfaces so that they can rock on one another, as 
shown in Fig. 151. The points of contact of successive voussoirs are 
then called centers of pressure, and the line joining them the line of 
pressure, or linear arch. It is evident, from the figure, or from a model 
constructed as above, that with every change of loading the voussoirs 
change their position more or less, thus altering the form of the 
linear arch. In a model constructed as above, the linear arch can 
alter its shape consider- 
ably without overthrow- 
ing the structure, the only 
condition necessary to 
assure stability being 
that the linear arch shall 
lie within the middle 
third.* 

In a masonry arch the 
pressure on any joint 
is ordinarily distributed 
over the entire surfaces in contact. In this case the center of pres- 
sure is the point of application of the resultant joint pressure, and 




* See discussion of arches in article by Fleeming Jenkin, entitled " Bridges," Ency- 
clopedia Britannica, 9th ed., Vol. IV, pp. 273-282. 



' AECHES AND AECHED KIBS 



219 



the linear arch is the broken line joining these centers of pressure. 
In a concrete arch the linear arch becomes a continuous curve. With 
each change of loading the same shifting of the linear arch occurs 
as in the case of the model with curved joints, the only difference 
being that with flat joints this action is not visible. To assure sta- 
bility, however, the linear arch must be restricted to lie within the 
middle third of the arch ring, as will be proved in Article 148. 

If we consider a single voussoir of a masonry arch, or a portion of 
a concrete arch bounded by two plane sections, as shown in Fig. 152, 
the resultant joint pressures R and R', and the weight P of the 
block and the material directly above it, form a system of forces in 
equilibrium. Consequently, if the amount, direction, and point of 
application of one of these 
resultant joint pressures 
are known, the amount, 
direction, and point of ap- 
plication of the other can 
be found by construct- 
ing a triangle of forces. 
Therefore, if one result- 
ant joint pressure is com- 
pletely known in position, 
amount, and direction, the others can be successively found as above, 
thus determining the linear arch as an equilibrium polygon for the 
given system of loads. 

Since an equilibrium polygon may be drawn to any given scale, 
if no one joint pressure is completely known, which is usually the 
case, there will be, in general, an infinite number of equilibrium 
polygons corresponding to any given system of loads. The linear 
arch may, however, be defined as that particular equilibrium polygon 
which coincides with the pressure line, and the question then arises 
how to determine the equilibrium polygon so that it shall coincide 
with the pressure line. This problem will be discussed more fully in 
Articles 150, 151, and 152. 

When the linear arch has been determined, the resultant pressure 
on a joint having any inclination to the vertical can easily be 
obtained. Thus, in Fig. 153, let R be the resultant pressure on a 




220 



STRENGTH OF MATERIALS 



DC 




R' 



FIG. 153 



vertical section through B, and R' the resultant pressure on the 
inclined section AE through B. Since R is due to the load on the 
right of the vertical CF, and R 1 to the load 011 the right of the broken 
line DAE, the difference between them must be due to the load 

ABCD minus the load BFE. Let 
P denote the difference between 
these two loads, represented by 
the shaded portion in Fig. 153. 
Then, since R, R J , and P must 
be in equilibrium, R' is found at 
once by drawing a force triangle, 
as shown in the figure. 

148. Conditions for stability. 
A masonry arch may fail in any 
one of three ways : (1) by sliding 
of one voussoir upon another; (2) by overturning; (3) by crushing 
of the material. 

These three methods of failure will now be considered in order. 

1. The first method of failure is caused by the shearing stress at 
any joint exceeding the joint friction, or the adhesion of the mortar. 
This kind of failure can only occur when the angle which the result- 
ant pressure on any joint makes with a normal to the plane of 
the joint exceeds the angle of repose for the material in question 
(Article 167). Ordinarily the resultant pressure on any joint is very 
nearly perpendicular to its plane, and since the angle of repose for 
masonry is very large, failure by sliding is not likely to occur. 

As a criterion for safety against failure of this kind, it may be 
assumed that when the resultant makes an angle of less than 30 
with the normal to the joint safety against sliding is assured. 

2. In order for an arch to fail by overturning, one or more of the 
joints must open at one edge, the adjacent blocks rotating about 
their center of pressure. For this to occur, one edge of the joint 
must be in tension. Although in a well-laid masonry arch the 
joints have considerable tensile strength, it is customary to disregard 
this entirely, and in this case the condition necessary to assure 
stability against rotation is that every joint shall be subjected to 
compressive stress only. Assuming, then, a linear distribution of 



ARCHES AND ARCHED RIBS 



221 



stress over the joints, the center of pressure is restricted to lie within 
the middle third of any joint (compare Article 62). 

Thus, in Fig. 154 (A), if ABCD represents the distribution of 
pressure on any joint AD, the resultant R must pass through the 
center of gravity of the trapezoid ABCD. Consequently, when the 
compression at one edge becomes zero, as shown in Fig. 154 (B), 
the resultant R is applied at a point dis- 

tant from A y and cannot approach any 

o 

nearer to A without producing tensile 
stress at D. Therefore, the criterion for 
stability against overturning is that the 
center of pressure on any joint shall not 

approach nearer to either edge than - > 

o 

where b is the width of the joint ; or, in 
other words, that the linear arch must lie 
within the middle third of the arch ring. 

3. Failure by crushing can only occur 
when the maximum stress on any joint 
exceeds the ultimate compressive strength 
of the material. To guard against this 
kind of failure, 10 is universally chosen 
as the factor of safety. Hence, if u c denotes the ultimate compressive 
strength of the material, and j9 ma ' x the maximum unit stress on any 
joint, the criterion for stability against crushing is 




FIG. 154 



From Fig. 154 (B), the maximum unit stress is twice the average. 
Therefore, if F denotes the area of a joint, and p a the average unit 
stress on it, P 



Consequently the criterion for stability against crushing can be 
expressed in the more convenient form 



F 20' 



222 STRENGTH OF MATERIALS 

that is to say, the average unit stress on any joint must not exceed 
one twentieth of the ultimate compressive strength of the material. 

The above conditions for stability can be applied equally as well 
to a concrete arch by considering the stress on any plane section of 
the arch ring. 

149. Maximum compressive stress. Let R denote the resultant 
pressure on any joint, b the width of the joint, F its area, and c the 
distance of the center of pressure from the center of gravity of the 
joint. Then, under the assumption of a linear distribution of stress, 
the stress on the joint is due to a uniformly distributed thrust of 

T> 

amount per unit of area, and a moment M of amount M = Re. 
Therefore the unit stress p at any point is given by the formula 

R .Me 

p = , 
r F I 

where e is the distance of the extreme fiber from the center of gravity, 
and / is the moment of inertia of the cross section. 

For a section of unit length, F = b-l =b, I = > and e = - . 

I u 

Therefore, substituting these values, the formula for maximum or 
minimum stress becomes 

R , 6 Re 

Pmax = ' 

min 

For e = - the minimum stress is zero, and if c > - it becomes nega- 
6 .6 

tive, thus restricting the center of pressure to lie within the middle 
third of the cross section if tensile stress is prohibited (compare 
Article 62 and Article 148, 2). 

Combining this result with that of the preceding article, the maxi- 
mum stress calculated by the formula 

_R 6 Re 

must not exceed - > where u c is the ultimate compressive strength 
of the material. 

150. Location of the linear arch : Moseley's theory. In order to 
obtain a starting point for the construction of the linear arch, it is 
necessary to know the amount, direction, and point of application 



AKCHES AND AKCHED KIBS 223 

of one joint pressure, as explained in Article 147 ; or, in general, it 
is necessary to have given three conditions which the equilibrium 
polygon must satisfy, such, for instance, as three points through 
which it is required to pass. Since it is impossible to determine these 
three unknowns by the principles of mechanics, the theory of the 
arch has long been a subject of controversy among engineers and 
mathematicians. 

Among the various theories of the arch which have been proposed 
from time to time, the first and most important of the older theories 
is called the principle of least resistance. This theory was introduced by 
the English engineer, Moseley, in 1837, and later became famous on the 
Continent through a German translation of Moseley's work by Scheffler. 

In building an arch the material is assembled upon a wooden frame- 
work called a center ; when the arch is complete this center is removed 
and the arch becomes self-supporting, as explained in Article 145. 
Now suppose that instead of removing the center suddenly, it is 
gradually lowered so that the arch becomes self-supporting by degrees. 
In this case the horizontal pressure or thrust at the crown gradually 
increases until the center has been completely removed, when it has 
its least possible value. This hypothesis of least crown thrust con- 
sistent with stability is Moseley's principle of least resistance. 

In constructing an equilibrium polygon the horizontal force, or 
pole distance, is least when the height of the polygon is a maximum. 
Therefore, in order to apply the principle of least resistance, the equi- 
librium polygon must pass through the highest point of the extrados 
at the crown and the lowest points of the intrados at the abutments. 
Since this would cause tensile stress at both the crown and abut- 
ments, the criterion for stability against overturning makes it neces- 
sary in applying the theory to move the center and ends of the 
equilibrium polygon, or linear arch, until it falls within the middle 
third of the arch ring. There is nothing in the principle of least 
resistance, however, to warrant this change in the position of the 
equilibrium polygon, and consequently the theory is inconsistent with 
its application. 

Culmann tried to overcome this objection to Moseley's theory by 
considering the compressibility of the mortar between the joints. At 
the points of greatest pressure the mortar will be compressed more 



224 STRENGTH OF MATERIALS 

than elsewhere, and this will cause the pressure line, or linear arch, 
to move down somewhat, thus taking a position nearer to the middle 
third than is required by the principle of least resistance, if applied 
to the arch as a rigid body. 

The above brief account of Moseley's principle of least resistance 
and Culmann's modification of it are given chiefly for their historical 
interest and the importance formerly attached to them. The modem 
theory of the arch is based upon the principle of least work, and is 
therefore rigorously correct from the standpoint of the mathematical 
theory of elasticity. 

151. Application of the principle of least work. Although Hooke's 
law is not rigorously true for such materials as stone, cement, and 
concrete, the best approximation to actual results is obtained by 
assuming that the materials of which the arch is composed conform 
to Hooke's law, and then basing the theory of the arch on the general 
theorems of the strength of materials. On this assumption the posi- 
tion of the linear arch can be determined by means of Castigliano's 
theorem, which states that for stable equilibrium the work of defor- 
mation must be a minimum (Articles 79 and 81). 

Consider a section of the arch perpendicular to the center line of 
the arch ring, or, in general, normal to the intrados. Let F denote 
the area of the section, R the resultant pressure on the section, c the 
distance of the point of application of R from the center of gravity 
of the section, and ds an infinitesimal element of the center line. 
Then the work of deformation will consist of two parts, that due 
to the axial thrust It, and that due to a moment M = Re. Since the 

T-> 

direct stress per unit of area of the section is > the unit deformation 

R 

due to the stress is > where E denotes Young's modulus; and 

FE -. / 7? \ 7? 2 

hence the work of deformation due to R is -R{ )> or 

2 \FE I 2 FE 

From Article 73, Chapter IV, the work of deformation due to the 

M 2 

bending moment M is Therefore the work of deformation dW 

2 El 

for a portion of the arch included between two cross sections at a 
distance ds apart is 

, T/r . R 2 , M* 

d W = - ds H ds. 

2EF 2 El 



ARCHES AND ARCHED RIBS 225 

Hence the total work of deformation for the entire arch is 

/R 2 C M 2 

2 EF d + J ^EI d 

Let b denote the thickness of the arch ring, and consider a section of 
unit width. Then F = b and / = > and substituting these values 

in the above equation and assuming that E is constant throughout 

the arch, -. 

W= 



In Article 147 it was shown that three conditions are necessary 
for the determination of the linear arch. Therefore, since the values 
of R and M in the above expression depend upon the position of the 
linear arch, in order to apply Castigliano's theorem to the integral, 
R and M must first be expressed in terms of these three unknown 
quantities, which may be conveniently chosen as the position, amount, 
and direction of the joint pressure at a certain point. 

Having expressed R and M in this way, Castigliano's theorem is 
applied by differentiating W partially with respect to each of the three 
unknowns, and equating these three partial derivatives to zero. In 
this way three simultaneous equations are obtained which may be 
solved for the three unknown quantities, thus completely determining 
the linear arch. 

The principle of least work, therefore, permits of a rigorously cor- 
rect determination of the linear arch. Instead, however, of actually 
carrying out the process outlined above, Winkler has applied the prin- 
ciple to the derivation of a simple criterion for stability, as explained 
in the following article. 

152. Winkler's criterion for stability. From the preceding article, 
the total work of deformation for the whole arch is given by the 
expression -, 



in which the integral is to be extended over the entire length of the 
arch. As the position of the pressure line is altered, the first term 
in this integral changes but little, whereas the second term under- 
goes a considerable variation, since M = Re, where c is the distance 



226 STRENGTH OF MATERIALS 

of the center of pressure from the center of gravity of the section. 
For a first approximation, therefore, the first term may be disregarded 
in comparison with the second, and hence the problem of making W 

/M 2 
ds as small 



as 

To effect a still further reduction, suppose that R is resolved into 
vertical and horizontal components so that the vertical component 
shall pass through the center of gravity G of the section (Fig. 155), 
and let z denote the perpendicular distance of the horizontal com- 

ponent P h from G. Then M = P h z and 




/jf 2 rp~s? 

ds becomes I ^- 



ds, 



b 

or, since P h is constant for all sections, 
this may be written P\ I - 

Ordinarily the thickness of the arch 
ring varies, being least at the crown 
FlG 165 and greatest at the abutments. In this 

case let b c denote the thickness of the 

crown, and suppose that the law of variation in thickness is such 
that the thickness I at any other point is given by the expression 

c ~dx 

where dx is the horizontal projection of ds. Under this assumption, 
the expression P\ I - becomes 



Therefore the problem of making W a minimum is now reduced to 
that of making the integral / z*dx as small as possible. 

This latter expression, however, consists of only positive terms, 
and reduces to zero for the center line of the arch. From this it 
follows that if an equilibrium polygon is drawn for the given system 
of loads, and then the center line of the arch is so chosen as to coin- 
cide with this equilibrium polygon, the true linear arch can differ 
but little from this center line. 



AECHES AND ARCHED KIBS 227 

In order for an arch to be stable at least one of the many possible 
assumptions of the linear arch must be such as to fall within the 
middle third of the arch ring. Moreover, the elastic deformation of 
the arch is such as to move the linear arch as near to the center line 
as the form of the arch permits. Therefore, if for any given arch 
it is possible to draw an equilibrium polygon which shall everywhere 
lie within the middle third of the arch ring, the stability of the arch 
is assured. 

This criterion for stability is due to Winkler, and was first given 
by him in 1879. 

153. Empirical formulas. The thickness necessary to give an arch 
at the crown can only be found by assuming a certain thickness and 
determining whether or not this satisfies all the conditions of sta- 
bility. The least thickness consistent with stability is such that the 
average compressive stress does not exceed one twentieth of the 
ultimate compressive strength of the material. The arch is usually 
made somewhat thicker than is required by this criterion, however, 
for the thicker the arch the more easily can the equilibrium polygon 
be made to lie within the middle third of the arch ring. 

The following empirical formulas for thickness at crown represent 
the best American, English, and French practice respectively, and 
may be used in making a first assumption as a basis for calculations. 



Trautwine. 
Rankine. 




r = radius of intrados in feet ; d rise in feet ; 

/ = span in feet ; b = depth at crown in feet. 

154. Designing of arches. In designing an arch to support a given 
loading the equilibrium polygon for the given system of loads should, 
in accordance with Winkler's criterion, be assumed as the center line 
of the arch. This, however, is not always possible, For instance, in 



228 STRENGTH OF MATERIALS 

the case of an arch intended to support a roadway, the level of which 
is fixed, the loading depends to a large extent on the form of the 
arch, and consequently the equilibrium polygon cannot be determined 
until the form of the arch has been assumed. 

In designing arches, therefore, the method usually followed is to 
assume the form of the intrados of the required arch, and determine 
its thickness at the crown by an empirical formula, such as those 
given in the preceding article. Then, having draw^n the extrados and 
load line, the surface between the intrados and the load line is 
divided into any convenient number of parts by drawing verticals, 
and the amount and position of the resultant weight of each part for 
a section one foot wide is calculated. An equilibrium polygon for 
this system of loads is then passed through the middle point of the 
arch ring at crown and abutments by the method given in Article 142. 
If this equilibrium polygon lies within the middle third of the arch 
ring, the arch is assumed to be stable against overturning. 

If the equilibrium polygon through the middle points of the arch 
ring at crown and abutments does not lie entirely within the mid- 
dle third of the arch ring, these three points are shifted so as 
to make it do so if possible. If no choice of the three points will 
make the equilibrium polygon lie entirely within the middle third 
of the arch ring, the design must be altered until this has been 
accomplished. 

The next step is to calculate the maximum unit joint pressure by 
the formula given in Article 149, and apply the criterion for stability 
against crushing given in Article 148. When these criteria have been 
satisfied the design is assumed to be safe. If, however, there is a 
considerable excess of strength, the design may be lightened and the 
criteria reapplied. 

Before the design can be considered complete it must also be 
shown that the above criteria are satisfied for every form of loading 
to which the arch is likely to be subjected. In the case of an arch 
designed to carry a heavy live load, such as that due to several 
locomotives, it may be necessary to draw a number of load lines 
corresponding to different positions of the load, and make a corre- 
sponding number of determinations of the equilibrium polygon and 
maximum joint pressure. 



AKCHES AND AKCHED KIBS 229 

The stability of the abutments still remains to be investigated, and 
finally the bearing power of the soil on which these abutments rest. 

Problem 288. Design a concrete arch to span a stream 25 ft. in width and sup- 
port a roadway 15 ft. above the level of the stream, if the spandrel filling is clay 
weighing 120 lb./ft. 3 ; the maximum depth of frost is 3 ft. and the bearing power 
of the soil at this depth is 4 tons /ft. 2 (see Article 158). 

155. Stability of abutments. To determine the stability of the 
abutments, the joint pressure at the haunch is combined with the 
weight of the abutment into a single resultant, say R 1 . For stability 
against overturning, the line of action of this resultant must strike 
within the middle third of the base (Article 148, 2). 

Eesolving the resultant R' into a horizontal component R h and a 
vertical component R v) the maximum pressure on the soil is calcu- 
lated by substituting this value of R v for R in the formula given in 
Article 149. To prevent sinking of the abutments, this pressure must 
not exceed the bearing power of the soil (see Article 166). 

For stability against sliding, the shearing stress between the abut- 
ment and the soil, due to the horizontal component R h of the result- 
ant R', must be less than the friction between the two; or, more 
briefly, the angle which R f makes with the horizontal must be less 
than the angle of repose (compare Article 172). 

156. Oblique projection of arch. Suppose that an arch, its load line, 
and its pressure line are drawn to any given scale, and then the whole 
figure is projected upon an oblique plane by a system of parallel lines. 
The projection of the pressure line on this oblique plane will then be 
the true pressure line for the projected arch and its projected load line. 

This principle can often be used to advantage, as, for example, in 
comparing two arches of equal span but different rise. Its most 
important application is in giving an accurate construction of the 
pressure line for arches of long span and small rise. Thus, instead of 
plotting such an arch to scale, its projection can be plotted ; or, in 
other words, its span can be shortened any convenient amount. A 
larger unit can then be used in plotting the vertical dimensions than 
would otherwise be possible, and consequently the pressure line can 
be drawn to any desired degree of accuracy. 

Having constructed the pressure line in this way, the pressure on 
any joint of the given arch can be found from the pressure on the 



230 



STRENGTH OF MATERIALS 



corresponding joint of the projected arch by laying off the horizontal 
and vertical components of the latter to two different scales ; in other 
words, by projecting the pressure back again onto the original arch. 



III. ARCHED RIBS 

*157. Stress in arched ribs. The arch is frequently used in metal 
constructions, especially in such structures as roofs and bridges, in 
the form of a curved beam composed either of a solid web and flanges 
or built up like a truss. Such a metal arch is called an arched rib. 




FIG. 156 

The fundamental difference between a concrete or masonry arch 
and an arched rib is that the latter, being composed of metal, is 
capable of resisting bending moment. For an arched rib, therefore, 
it is not essential that the equilibrium polygon shall lie within the 
boundaries of the arch ; it may, in fact, either cross the arch or lie 
entirely on either side, the only condition for stability being that the 
arched rib must be sufficiently strong to resist the bending moment 
thus produced. 

*For a brief course the remainder of this chapter may be omitted. 



AKCHES AND AKCHED KIBS 231 

When the equilibrium polygon has been drawn for the given system 
of loads, the stress at any point of an arched rib can be calculated by 
the method explained in Article 143. Thus, in Fig. 156 (^t), let AGF 
denote the arched rib, P v P 2 , etc. the given loads, and ABCDEFthe 
corresponding equilibrium polygon. Then the stress on any section 
mn is due to a force acting in the direction CD, of amount equal to 
the corresponding ray OC r of the force diagram. 

Consequently, if the rib is composed of a solid web and flanges, 
as shown in Fig. 156 (B), the direct stress on the section is equal in 
amount to the ray OC f of the force diagram, the bending stress on 

P z 1 P z 

the upper flange is ~ > the bending stress on the lower flange is - , 

Cu CL 

and the shear normal to the rib is OC f sin a, where a is the angle 
between CD and the tangent to the rib at the section. 

Similarly, for the trussed rib shown in Fig. 156 (C), by taking 
moments about L and S the stresses in ItS and LK are found to be 

P z P z' 

f- and 7- respectively, while the normal component of the stress 
a d 

in LS is OC sin a. 

Arched ribs are usually constructed in one of three different ways : 
(1) hinged at the abutments and at the crown; (2) hinged at the 
abutments and continuous throughout; (3) fixed at the abutments 
and continuous throughout. The method of constructing the equilib- 
rium polygon differs for each of these three methods of support, and 
will be treated separately in what follows. 

158. Three-hinged arched rib. When a member is free to turn 
at any point the bending moment at that point is zero, and con- 
sequently the equilibrium polygon, or bending moment diagram, 
passes through the point. For a three-hinged arched rib, therefore, 
the equilibrium polygon must pass through the centers of the 
three hinges and is therefore completely determined, as explained in 
Article 142. 

159. Two-hinged arched rib. Consider an arched rib hinged at 
the ends and continuous between these points. In this case the 
equilibrium polygon must pass through the centers of both hinges, 
but since there is no restriction on the vertical scale, this scale may 
be anything whatever, depending on the choice of the pole in the 



232 



STRENGTH OF MATERIALS 



force diagram. A third condition is therefore necessary in order to 
make the problem determinate. 

The problem can be solved in various ways, depending on the 
choice of the third condition. The first solution that will be given is 
that found by applying the principle of least work, that is, by apply- 
ing Castigliano's condition that the work of deformation shall be a 
minimum. 

Consider a two-hinged arched rib supporting a system of vertical 
loads, as shown in Fig. 157. Then the moment at any point A is 
equal to the moment of the forces on the left of the section mn 
through A, minus the moment of P h about A, where P h is the unknown 



A 



V 



FIG. 157 



horizontal reaction, or pole distance of the force diagram, which is to 
be determined. Consequently, if M denotes the moment at A, M p the 
moment of the forces on the left of A, and z the perpendicular distance 

of P h from A, we have 

M= M p - P h z. 

Since the work of deformation due to the shear and axial load is small, 
it may be neglected in comparison with that due to the bending mo- 
ment. Under this assumption the work of deformation is 



El 



ds, 



in which the integral is to be extended over the entire length of 
the rib. Applying the principle of least work to this expression, the 
partial derivative of W with respect to the unknown quantity P h 
must be zero. Hence 



ARCHES AND ARCHED RIBS 



233 



or 



whence 



T> 



/M p z 



/-* 

If E is constant throughout the rib, this reduces to 

j-fds 



P h = 



The pole distance P h found from this formula is the third condition 
necessary for the complete determination of the equilibrium polygon. 

160. Second method of calculating the pole distance. The value 
of the pole distance P h of the force diagram can also be calculated by 
assuming that the bending of the rib produces no change in the span. 
To apply this condition, the change in length which the span would 
naturally undergo is 
calculated and equated 
to zero. 

Consider a small 
portion ds of the rib. 
If, for the moment, 
the rest of the rib is 
regarded as rigid, the 

bending of this portion would make the end B revolve about D as 
a center to a position at C (Fig. 158). Let d/3 denote the angle 
between DB and DC, a the angle between DB and a vertical through 
D, z the ordinate DF, and CE, or A/, the change in length of the 
span. Then 

BC = DB dft, DB cos a = z, and A/ = CB cos a. 
Hence A/ - DB ' d/3 cos a = zd/3. 




FIG lgg 



234 



STRENGTH OF MATERIALS 



From Article 66, the angular deformation d/3 is given by the expression 

_ Mds 

Consequently 



and hence the total change in length of the span is * 

CMz 
1=1 ds. 



Therefore the condition that the span shall be unchanged in length 

by the strain is rw 

ds = Q. 
J EI 

The bending moment M in this expression has the same value as 
in the preceding article, namely, M = M p P h z. Inserting this value 

of If in the above condi- 
tion, it becomes 




EI 

from which, as in the pre- 
ceding article, 



P 



r*? 

j j 



j 

d s 



FIG. 159 



161. Graphical deter- 
mination of the linear 

arch. From the condition that the bending stress shall produce no 
change in the length of the span, the position of the linear arch may 
be determined graphically as follows. 

In Fig. 159 let ACF represent the center line of the rib, ADF the 
corresponding equilibrium polygon drawn to any convenient scale, and 
ABF the linear arch. Then the linear arch can be obtained from the 
equilibrium polygon by reducing each ordinate of the latter in a certain 

* The effects of changes of temperature and also of direct compressive stress in altering 
the length of the span are neglected, as they are slight in comparison with that due to 
bending strain. 



ARCHES AND ARCHED RIBS 235 

ratio, say r. The problem then is to find this ratio r in which the 
ordinates to the equilibrium polygon must be reduced to give the 
linear arch. 

The condition that the span is unchanged in length, derived in 
the preceding article, is 



hi which z represents the ordinate CE to the rib, and ds an element 
of the rib. Since the bending moment M is proportional to the verti- 
cal intercept between the linear arch and the center line of the rib, 
this condition may be written 

*BC-z 7 



f 



El 

or, since E may be assumed to be constant and 

BC = BE - CE = BE - z, 
this condition becomes 






which may be written 



If r denotes the ratio in which the ordinates to the equilibrium 
polygon must be decreased in order to give the linear arch, then 

BE 



r = - 
DE 



and consequently the condition becomes 

Cr-DE-z Cz* , 

J - ds- I jds=Q; 



whence 



DE-z 



f 

This expression for r can be evaluated graphically by replacing 
the integrals by summations and calculating the given functions for a 



236 STRENGTH OF MATERIALS 

series of vertical sections taken at equal intervals along the rib. Thus, 
since ds in this case is constant, 



r = 



I 

in which the functions under the summation signs are to be calculated 
for each section separately, and their sum taken. After r has been 
found in this way the linear arch is obtained by decreasing the 
ordinates of the equilibrium polygon in the ratio r : 1, and the stress 
can then be calculated as explained in Article 157. 

This method of determining the linear arch is due to Ewing. 

162. Temperature stresses in two-hinged arched rib. When the 
temperature of an arched rib changes, the length of the rib also changes, 
and consequently stresses called temperature stresses are produced in 
the rib (compare Article 19). To calculate the amount of this stress 
let L denote the coefficient of linear expansion and T the change in 
temperature in degrees. Then each element of the rib of length ds 
changes its length by the amount LTds, the horizontal projection of 
which is LTdx. Therefore the total change I in the length of the rib is 

1= C C LTdx = 2cLT, 

Jo 

where 2 c is the span. From Article 160, the total change in length 
of the span is given by the expression 



I 

Therefore 



_ f*Mz 
~Jo El 

f**-2 

Jo El 



cLT. 
.0 El 

To simplify this expression assume that the modulus of elasticity 
E is constant throughout the rib, and that the moment of inertia / 

ds 
increases towards the abutments in the ratio Under this assump- 

ds dx 

tion / = I - > where J denotes the moment of inertia at the crown, 
dx 

and the above equation becomes 
I 



El. 



Mzdx = 2 cLT, 



ARCHES AND ARCHED RIBS 



237 



The only forces which tend to resist the change in length of the rib 
due to temperature stresses are the horizontal reactions P h of the 
abutments. Therefore the external moment at any section of the rib 
with ordinate z is M = P h z, and substituting this value in the above 
integral, it becomes 



whence 



2 EI Q cLT 



z 2 dx 



This expression is easily evaluated in any given case, thus determin- 
ing P h and consequently the linear arch. The temperature stresses 
can then be calculated by the methods explained above, and combined 







>l 



FIG. 160 



with those due to the given loading. For a rise in temperature above 
that for which the arch was designed, T is positive and the horizontal 
reactions P h of the abutments act inwardly ; for a fall in temperature 
T is negative and the reactions P h act outwardly. 

To illustrate what precedes, the above formula will now be applied 
to a parabolic arched rib, which on account of its simplicity is the 
form ordinarily assumed in designing. Let h denote the rise of 
the arch, 2 c its span, and x, z the coordinates of any point A on 
the rib (Fig. 160). Then, from the intrinsic property of the parabola, 

it follows that 

h z (c x) . 

"T" ~~s~ 

whence 

fix ._ , 

z = (2 c - x). 
c 



238 STRENGTH OF MATERIALS 

/c 
z z dx and integrating, 



we have 



f 2 

Jf 



Consequently for a parabolic arched rib the horizontal reaction of 
the abutments due to a change in temperature of T degrees is 

^A = 



163. Continuous arched rib fixed at both ends. For a continuous 
arched rib fixed at both ends the problem of constructing the equi- 
librium polygon is subject to a threefold indetermination, since none 
of the three conditions necessary for its determination are given. 
The theoretical solution of the question by the principle of least 
work is as follows. 

Let the vertical reaction R^ t the horizontal reaction P h , and the 
bending moment M at the left support be chosen as the three 
unknown quantities necessary to determine the linear arch. For a 
system of concentrated loads the moment M at any section of the 
rib distant x from the left support is 



M 



= M + Ex - P h z - V P(x - d), 



in which d is the distance of any load P from the left support, and 
the summation is to be extended over all the loads between the left 
support and the point under consideration. Similarly, for a uniform 
load of amount w per unit of length, 



Now, from Article 73, the work of deformation W is given by the 
expression 



in which M has the value given by one or the other of the above 
expressions, depending on whether the loading is concentrated or 
uniform. To apply Castigliano's theorem to this expression it is 



ARCHES AND ARCHED RIBS 



239 



necessary to find the partial derivatives of W with respect to M , R v 
and P h respectively, and equate these derivatives to zero. The three 
conditions obtained in this way are 



dW 



M dM 



Mx 



(103) ^= ^7-^ ds = 



M 



since from either of the above expressions for M we have 

= x, and - = z. Inserting in these three conditions the value 

h 

of M for the given form of loading, three simultaneous equations are 
obtained which may be solved for the three unknown quantities R v 
P h , and M . 

Equations (103) can also be obtained by assuming as our three con- 
ditions that the horizontal and vertical deflections of the supports are 
zero, and that the direction of the rib at the ends remains unchanged. 
The method of obtain- D 

ing equations (103) from 
these assumptions is 
simply an extension of 
that given in Article 
160 for the two-hinged 
arched rib. 

164. Graphical deter- 
mination of the linear 
arch for continuous 
arched rib. The simplest 
method of applying equa- 
tions (103) to the deter- 
mination of the linear 
arch is by means of a graphical treatment similar to that given in 
Article 161. 

Consider first the case of symmetrical loading. Then if M denotes 
the bending moment at either abutment, the linear arch has the same 




E' 



E 
FIG. 161 



240 STRENGTH OF MATERIALS 

form as for a rib with two hinges, except that its base is shoved 
down a distance M below the springing line of the rib. Therefore 
in this case the linear arch is completely determined by the two 
quantities M and r, the third condition being supplied by the sym- 
metry of the figure. 

In Fig. 161 let ACF represent the center line of the rib, A'BF' 
the linear arch, and ADF the equilibrium polygon for the given 
system of loads. Since the bending moment M at any point of the 
rib is the vertical intercept BC between the linear arch and the 
center line of the rib, we have 

M=BC = BE-CE* -EE', 
or, since BE = r-DE', M=r . DE > - Z -M V 

Substituting this value of M in the first and third of equations (103), 
they become 



CrDE'zds_r j ?_ TM^ 

J El J El J El 

If the expressions under these integral signs are evaluated for a num- 
ber of vertical sections taken at equal distances along the rib, and the 
results are summed, we obtain the two conditions 






from which r and M can easily be determined. The linear arch is 
then constructed by starting from a point at a distance M below the 
left support, and decreasing the ordinates to the equilibrium polygon 
in the ratio r : 1. 

If the loading is unsymmetrical, the moments at the ends of the 
rib are not equal. Let M l and M 2 denote the moments at the left 
and right ends respectively (Fig. 162). As before, the moment M 
at any point of the rib is the vertical intercept BC between the linear 
arch A'BF' and the center line of the rib ACF. Consequently 

M = BC = BE - CE' - EE'. 



ARCHES AND ARCHED RIBS 



241 



In this case, however, the distance EE* is not constant from A to F, 
but varies as the ordinates to a triangle, being equal to M l at A and 
to M 2 at F. Hence, for a point at a distance x from A, 



(EE') X = M l - - (M, - 



where 2 c is the length of the span. Also BE = r - DE, and CIS' = z. 
Therefore 

M = r DE - z - M, -f - (M. - M z ). 
2 c 

Let this value of M be inserted in equations (103). Then, if the 
expressions under the integral signs are evaluated for a number of 




vertical sections taken at equal distances along the center line of the 
rib, and their sums taken, the integrations in equations (103) can be 
replaced by summations giving the three conditions 



DE ^ z ^ f \^ 1 

-T-L-I- M ^LJ 



zx 



ZDE-z 
-- 



2c 



242 STEENGTH OF MATEEIALS 

Solving these three equations simultaneously for M lt M 2 , and r, the 
linear arch is constructed by laying off M l and M 2 from A and F 
respectively, and then reducing the ordinates to the equilibrium 
polygon in the ratio r : 1, and laying them off from the line A'F'. 

The stresses in the rib can then be calculated by the methods 
previously given (Article 157). 

165. Temperature stresses in continuous arched rib. Using the 
notation of Article 162, the change in the length of the span due to 
a change in temperature of T degrees is 

/ = 2 cLT. 



Therefore, for temperature stresses equations (103) become 

Mz 
~EI 



MX , r J 

zi s ~~ ' J * 



By hypothesis, the only external forces acting on the rib are the 
reactions and moments at the abutments due to the temperature 
stresses. Consequently, if R denotes the vertical reaction, P h the hori- 
zontal reaction, and M^ the moment at the left abutment, the moment 
M at any other point of the rib is 

M= M^ + Rx- P h z. 

If, then, this value of M is inserted in the above integrals and the 
resulting equations solved simultaneously for M lt R, and P h) the linear 
arch is thereby determined. 



CHAPTER XI 

FOUNDATIONS AND RETAINING WALLS* 

166. Bearing power of soils. Since the character of a foundation 
is dependent upon the nature of the soil on which it is to rest, it is 
necessary in designing a foundation to know with a reasonable degree 
of accuracy the maximum load which the soil can sustain per unit 
of area without appreciable settlement; or, in other words, what is 
known as the bearing power of the soiLf 

Ordinarily the results of previous experience are relied upon to 
give an approximate value of the bearing power of any given soil, 
and stability is assured by the adoption of a large factor of safety. 
For structures of unusual importance, however, or when the nature 
of the soil is uncertain, the results of previous experience are usually 
insufficient to assure stability, and special tests are necessary for the 
determination of the bearing power of the soil in question. Among 
notable structures for which such special tests have been made may 
be mentioned the State Capitol at Albany, N.Y. ; the Congressional 
Library at Washington, D. C. ; the suspension bridges at Brooklyn, N. Y., 
and at Cincinnati, Ohio; the Washington Monument; the Tower 
Bridge, London, etc. 

By averaging the results of a large number of such tests, reliable 
information is furnished as to the bearing power of soils in general. 
The most commonly accepted of such average values are those given 
by Professor I. O. Baker in his Treatise on Masonry Construction, and 
are as shown in the table on the following page. Other values in 
common use are also quoted for comparison, and may be accepted as 
representative of modern practice. 

* For a more detailed treatment of foundations and retaining walls the following 
special treatises may be consulted. Baker, Treatise on Masonry Construction ; Howe, 
Retaining Walls for Earth; Fowler, Ordinary Foundations; Merriman, Walls and 
Dams; Patton, Ordinary Foundations. 

t The bearing power of soils is analogous to what is called the crushing strength in 
the case of more rigid materials, such as stone and brick. 

243 



244 



STRENGTH OF MATERIALS 



MATERIAL, 


BEARING POWER 
tons/ft.' 


Rock equal to best ashlar masonry 
Rock equal to best brick masonry 
Rock equal to poor brick masonry 


25 
15 
5 


Dry clay 


4 


Moderately dry clay .... 


2 


Soft clay . ... . . 


1 


Cemented gravel and coarse sand 
Compact and well-cemented sand 
Clean dry sand ... 


8 
4 

2 


Quicksand and alluvial soils . 


i 







As an approximate working rule Trautwine recommends from 2 
to 3 tons/ft. 2 as a safe load for compact gravel, sand, or loam, and 
from 4 to 6 tons/ft. 2 if a few inches of settlement may be allowed.* 

The building laws of Greater New York may also be regarded as 
competent authority, and specify the following values. 



MATERIAL 


BEARING POWER 

tons /ft .2 


Firm, coarse sand, stiff gravel, or hard clay 
L/oam clay or fine sand firm and dry 


4 
3 


Ordinary clay and sand together, wet and springy . . . 
Soft clay .... 


2 
1 







As a supplement to the above, these laws also specify that when 
foundations are carried down through earth by piers of stone or 
brick, or by concrete in caissons, the loads on same shall not exceed 
15 tons/ft. 2 when carried down to rock, or 10 tons/ft. 2 when carried 
down to firm gravel or hard clay. 

In order to obviate too large or expensive a foundation, it is often 
desirable to increase the bearing power of the soil. This may be 
accomplished in various ways. 

Since, in general, soils are more condensed at greater depths, 
increasing the depth usually increases the bearing power of the soil. 

* Engineer's Pocket-Book, 1902, p. 583. 



FOUNDATIONS AND RETAINING WALLS 



245 



In the case of wet or moist soils the same effect is obtained by 
drainage, as indicated in the tables on the preceding page. 

A more marked increase in the bearing power may be obtained by 
excavating the soil and replacing it by a layer of moist sand ; or by 
driving short piles and then either removing them and filling the 
hole immediately with moist sand, or else leaving the piles in the 
earth and covering them with a platform of timber or concrete. 

When none of these methods will suffice, the soil must be exca- 
vated until a subsoil with an adequate bearing power is reached. 

167. Angle of repose and coefficient of friction. When a mass of 
granular material, such as sand, gravel, or loose earth, is poured upon 
a level surface, the sides of the pile will assume a definite slope, 
called the natural slope. This maximum angle which the sides of 
the pile can be made to assume with the horizontal is called the 
angle of repose, and is a constant for any given material. Since the 
size of this angle is dependent upon the amount of friction between 
the particles of the 
material, it may be 
taken as a measure 
of the friction, "or 
vice versa. 

The laws of fric- 
tion as determined 

by experiment are 
/ 
that the force of 

friction is independent of the areas in contact, is dependent on the 
nature of the material, and is directly proportional to the normal 
pressure between the surfaces in contact. Let P F denote the force 
of friction and P N the normal pressure. Then the above laws may 
be expressed by the formula 

P- Z-P 
F ^-L N> 

where k is the constant of proportionality, and is called the coefficient 
of friction. 

In Fig. 163 let DE represent the natural slope and o> the angle 
of repose, and consider a particle of the material of weight P at any 
point A in the natural slope. Let P be resolved into two components 




FIG. 163 



246 



STKENGTH OF MATERIALS 



P F and Py, respectively parallel and perpendicular to DE. Then 
P F = P N tan a), and comparing this with the relation P F kP N > 

k = tan CD ; 

that is to say, the coefficient of friction is equal to the tangent of the 
angle of repose. 

The following table gives the numerical values of the angles of 
repose and coefficients of friction for various materials, and also the 
weight in pounds of one cubic foot of each material* 



MATERIAL, 


ANGLE OF 
REPOSE 


COEFFICIENT OF 
FRICTION 
k = tan to 


WEIGHT 
Ib./ft.s 


Sand, dry and fine 


28 


.532 


110 


" dry and coarse .... 
" moist .... 


30 

40 


.577 
839 


95 
110 


" wet 


30 


577 


125 


Clav damp . 


45 


1 000 


125 


" wet 


15 


268 


150 


Clayey gravel 


45 


1 000 


120 


Shingle . . . 


42 


900 




Gravel 


38 


781 


110 


Alluvial soil 


35 


700 


90 


Peat 


20 


364 


52 


Concrete best 






160 


" porous 






130 


Brickwork . .... 


33 


649 










140 


medium . ... 






125 


soft 






100 


Masonry 


31 


.601 




granite or limestone 






165 


sandstone 






144 


mortar rubble 






154 


dry rubble 






138 











168. Bearing power of piles. The custom of driving piles into the 
soil to increase its bearing power is of very ancient origin, and is still 
frequently used because of its cheapness and efficiency. Until quite 

* See Fanning, Treatise on Hydraulic and Water Supply Engineering, 15th ed., 1902, 
p. 345 ; Trautwine, Engineer's Pocket-Book, 1902, pp. 407-411 ; Smithsonian Physical 
Tables, 1896, Table 95; also the results compiled by Rankine from experiments by Gen- 
eral Morin and others, ibid., Table 149. 



FOUNDATIONS AND RETAINING WALLS 247 

recently wood was the only material used for piles, and they were 
either driven by hand with sledges, or by means of a block, usually 
of metal, which was raised between two upright guides and allowed 
to fall on the head of the pile. The latter form of pile driver is still 
in frequent use for driving wooden piles, and is called the drop- 
hammer pile driver. 

In 1839 Nasmyth invented the steam pile driver, which consists 
essentially of a steam cylinder supported vertically above the head 
of the pile by two uprights fastened to a cap which rests on the 
pile. The hammer in this case is a weight attached to the piston 
rod, and delivers a blow on the head of the pile at each stroke 
of the piston. The uprights which support the cylinder also serve 
as guides for the hammer, which varies in weight from 550 Ib. to 
4800 Ib. This form of pile driver owes its efficiency to the rapidity 
with which the blows can be given, the number being from sixty 
to eighty per minute, thus preventing the soil from recovering its 
equilibrium between strokes, and greatly decreasing its resistance to 
penetration. 

In modern engineering practice cast-iron and concrete piles are rap- 
idly coming into use, and as neither of these materials is capable 
of standing repeated blows, piles of this kind are usually driven by 
means of an hydraulic jet. The jet is attached to the point of the 
pile, thus constantly excavating the soil in front of the pile as it 
descends, and enabling it to sink into place with little or no assist- 
ance other than its own weight. 

The rational formulas in ordinary use for determining the bearing 
power of piles are based upon the assumption that the pile is driven 
by a drop-hammer pile driver, and express its bearing power in terms 
of the amount of penetration at the last blow. Since the bearing 
power of a pile is due in part to the friction of the earth on the sides 
of the pile, as well as to the resistance of the subsoil to penetration, 
and also since part of the energy of the hammer is absorbed by the 
friction of the guides, in compressing the head of the pile, in compress- 
ing the hammer, in overcoming the inertia of the pile, etc., a rigorous 
formula is too complicated to be of much practical value, although 
there are a number of elaborate discussions of the bearing power of 
piles which take all of these elements into consideration, notably the 



248 STRENGTH OF MATERIALS 

theories of Kankine and Weisbacli.* However, as several of the ele- 
ments entering into the discussion are attended with considerable 
uncertainty, it is customary in practice to use either an empirical 
formula or the simple approximate formula deduced below, adopting 
a factor of safety large enough to cover the assumptions made. 

Let P denote the weight of the hammer in pounds, h the height 
of the fall in inches, R the average resistance of the soil to penetra- 
tion during the last blow in pounds, and d the penetration of the 
pile, due to the last blow, in inches. Then, assuming that all the work 
done by the hammer is expended in overcoming the resistance of the 
earth at the point of the pile, we have 

Ph = Rd. 

With a factor of safety of 6, the approximate formula for safe load 
on the pile becomes 

(104) S = . 

As the head of a timber pile becomes " broomed " by repeated blows, 
and this greatly decreases the efficiency of the blow by absorbing the 
kinetic energy of the hammer, the head should be sawed off to a solid 
surface before making a test blow for determining the bearing power 
of the pile. 

For a drop-hammer pile driver the empirical formula in most 

common use is 

2Ph 

(105) K = dTl' 

the notation being the same as above, and the factor of safety being 6. 
For a steam pile driver this formula becomes 

<"> "iSn- 

where Ph represents the kinetic energy of the hammer. 

The above empirical formula, (105) or (106), is commonly known 
as Wellington's formula, or the Engineering News formula, and has been 
incorporated in the building laws of Greater New York. 

The only means of determining the bearing power of a pile driven 
by an hydraulic jet, is to observe the maximum load it can support 
without appreciable settlement. 

* See Baker, Treatise on Masonry Construction, chap. xi. 



FOUNDATIONS AND RETAINING WALLS 249 

Problem 289. A one-ton hammer falls 15 ft. on the head of a pile, and the 
settlement is observed to be . 1 in. Calculate the safe load for the pile by formulas 
(104) and (105) and compare the results. 

Problem 290. Under what conditions will the approximate rational formula 
(104) and the Engineering News formula (105) give substantially the same results? 

Solution. If the values of B obtained from these two formulas were equal, then 

= ; whence d Jj- in. For other values of d the rational formula gives the 

greater value of the bearing power when d < -fa in. , and the empirical formula gives 
the greater value when d > T J T in. From this it follows that the empirical formula 
is only applicable when the settlement at the last blow is small. 

169. Ordinary foundations. Although the foundation of a struc- 
ture is necessarily the first part to be constructed, it is the last part 
to be designed, for the weight of the structure determines the nature 
of the foundation, and this cannot be calculated until the structure 
has assumed definite proportions. 

The load which a structure is designed to carry consists primarily 
of three parts. 

1. The dead load, due to the weight of the structure and the per- 
manent fixtures, such as plumbing and heating apparatus, elevators, 
water tanks, machinery, etc. 

2. The live load, which depends on the use to which the structure 
is to be put, and which may vary from 20 lb./ft. 2 to 400 lb./ft. 2 

3. The wind load, due to the overturning action of the wind upon 
the side of the structure. These three parts of the total load must 
be calculated separately and then combined so as to give the maxi- 
mum resultant. The area of the foundation is then found at once 
by dividing this maximum load by the safe bearing power of the soil. 

The chief concern in designing a foundation, however, is not that 
its settlement shall be zero, but that it shall be uniform throughout. 
For if one part of a foundation settles more than another, it is evident 
that cracks are bound to occur which will seriously weaken the struc- 
ture and may even destroy its usefulness altogether. Since uniformity 
of settlement implies uniformity of pressure on the soil, the condition 
which determines the stability of a foundation and its superstructure 
is simply uniformity of pressure on the soil. 

The effect of violating this condition is frequently seen, the most 
common instance being that of ordinary dwelling houses in which 
several openings, say a door and a number of windows, occur one 



250 



STRENGTH OF MATERIALS 



above another. It is evident in this case that if the foundation is of 
the same width throughout, the centers of pressure will fall outside 
the centers of resistance, which will tend to throw the top of the wall 
outward on either side, and so result in cracks between the openings 
(Fig. 164). The remedy for this is either to narrow the foundation, 
or omit it altogether under the openings, or else extend it beyond 

the ends of the wall, the length of 
this extension being of such amount 
that the centers of pressure will fall 
inside, or at least coincide with, the 
centers of resistance. 

When a foundation extends 
beyond the ends of a wall the projec- 
tion is called the footing. To dimen- 
sion the footing it may be regarded 
as a simple cantilever, and its 
thickness calculated by the ordinary 
theory of beams. Thus let h denote 
the thickness of the footing in 
inches for a concrete foundation, 
or the thickness of the bottom foot- 
ing course in inches for a masonry 
foundation, b the width of the foot- 
ing in inches, u the ultimate strength of the material in lb./in. 2 , and 
P the load in tons/ft. 2 Then, since 1 ton/ft. 2 = 13.9 lb./in. 2 , the 
moment at the face of the wall is 




FIG. 164 



jr.* 

T bh * A 2 t, 41.7 , 

or, since / = and u = - > we have u = - - - ; whence 
2i I h 



fp 
h = 6.45 x -\] , approximately. 



Problem 291. Find the thickness of the bottom footing course for a masonry 
foundation if the load is 1 ton/ft. 2 , the factor of safety is 10, the footing is to 
extend 18 in. beyond the face of the wall, and is composed of limestone for which 
u = 15,000 lb./in.2 



FOUNDATIONS AND EETAINING WALLS 



251 



IIII 



CONCRETE 



170. Column footings. In the modern construction of tall build- 
ings the design frequently provides that the entire weight of the 
building and its contents shall be carried by a steel framework of 
columns and 
girders. This 
"skeleton type" 
of tall-building 
construction, as 
it is called, ne- 
cessitates a new 
type of founda- 
tion, since each 
column load 
must be calcu- 
lated separately 
and transmitted 
to the soil by a 
footing of suffi- 
cient size to 
give the neces- 
sary amount of 
bearing area. 

If the col- 
umns reach 
solid rock, the 
footing may 
consist simply 
of a base plate of 
such form as to 
give the column 
a solid bear- 
ing and afford 
sufficient anchorage to prevent the footing from lateral movement. 

For compressible soils the column is usually supported by a cast- 
iron base plate resting on a footing consisting of two or more layers 
of steel rails or I-beams, the whole resting on a concrete base, as 
shown in Fig. 165. 




J 



FIG. 165 



252 STRENGTH OF MATERIALS 

What has been said in the preceding article in regard to the cal- 
culation of the loads carried by the foundation also applies to the 
calculation of column loads, and the method of designing a column 
footing is essentially the same as for a masonry footing, explained 
above. Thus let P denote the total column load in tons, c the length 
of one side of the base plate in inches, and I the length in inches of 
the beams supporting it (Fig. 165). Then, if the base plate is assumed 
to be stiff enough to carry the load on its perimeter, the maximum 
moment M will occur at one edge of the base plate. Since the reac- 
tion on one side of the base plate is 2000 P--^> the amount of 

this moment is 

2000P(/-c) l-e 250P(/-c) 2 . 

~2l~ ~ ~T 

Consequently, if n is the number of beams supporting the base plate, 
the maximum moment for one beam is 

250P(/-c) 2 . 

M. = ^ ' in. Ib. 

nl 

If the base plate is assumed to be only stiff enough to distribute 
the load uniformly, the maximum moment will occur at the center 
of the beams, and its value will be (cf. Article 52 (E)) 



2000 P 



('-D. 



M= - i - =L = 250 P(2 I - c) in. Ib. 

In this case the maximum moment for one beam is 

250P(2J-f) . 

M, = - in. Ib. 

n 

Now let p denote the allowable fiber stress per square inch, / the 
moment of inertia of a cross section of one beam, and e half the 
depth of the beam. Then the moment of resistance of one beam is 



For foundation work p is usually taken to be 20,000 lb./in. 2 Substi- 
tuting this value, the moment of resistance becomes 

M= 20,000- = 20,000 S, 

e 



FOUNDATIONS AND RETAINING WALLS 253 

where S denotes the section modulus. Equating the moment of 
resistance to the external bending moment and solving the resulting 
equation for S, we have in the first case 






SO In 

and in the second case T> /*> 7 \ 

^ I C) 



In designing a column footing the column load P is first calculated, 
and the area of the footing determined by dividing the column load 
by the safe bearing power of the soil. The size of base plate and 
number of beams supporting it are next assumed, and the section 
modulus calculated by one of the above formulas. The size of beam to 
be used is then determined by choosing from the tables a beam whose 
section modulus agrees most closely with the calculated value of S. 

Problem 292. Design the footing for a column supporting~a load of 400 tons, 
and resting on a base plate 4 ft. square, so that the pressure on the foundation bed 
shall not exceed 3 tons/ft. 2 

171. Maximum earth pressure against retaining walls. A wall 
of concrete or masonry built to sustain a bank of earth, or other 
loose material, is called a retaining wall. 

In Chapter X it was shown that in order to determine the stability 
of an arch three conditions were necessary, which might conveniently 
be chosen as the direction, amount, and point of application of the 
resultant pressure on any cross section of the arch ring. The same 
necessity arises in the discussion of retaining walls, namely, that three 
conditions are necessary for the complete solution of the problem, 
and a number of theories have been advanced, notably those of 
Coulomb, Weyrauch, and Eankine, based on different assumptions as 
to these conditions. 

All theories, however, agree upon two of these assumptions, namely, 
(1) that the pressure against the wall is due to a wedge of earth, or, 
in other words, that the surface along which the earth tends to slide 
against the wall is a plane ; and (2) that the point of application of 
the resultant earth pressure is one third of the height of the wall 
from the bottom. Neither of these assumptions is rigorously correct, 
for the first is equivalent to neglecting the cohesion of the earth, and 



254 



STRENGTH OF MATERIALS 



the second assumes that the earth pressure against the wall is the 
same as if the earth was a liquid. However, the uncertainty attend- 
ing the exact degree of homogeneity of the materials under consid- 
eration probably does not warrant any greater precision in these first 
two assumptions. 

The third assumption relates to the direction of the maximum 
pressure, and is the point on which the various theories differ. Thus 
Coulomb and Weyrauch assume that the pressure is normal to the 

D 




FIG. 166 

back of the wall ; Rankine assumes that it makes an angle with the 
back of the wall equal to the angle of repose of the material ; while 
other authorities assume values intermediate between these two. 

In the present discussion the first two conditions mentioned above 
will be retained, and the third condition will be replaced by the 
assumption that the resultant earth pressure makes an unknown 
angle with a normal to the back of the wall. The assumptions 
are, then : 

1. The surface of rupture is a plane. 

2. The point of application of the resultant pressure is one third 
of the height of the wall from the bottom. 



FOUNDATIONS AND RETAINING WALLS 255 

3. The resultant pressure is inclined at an angle to a normal to 
the back of the wall. 

From the result of the theory based on these assumptions, the 
values of the resultant earth pressure given by Coulomb, Weyrauch, 
Kankine, and others will then be deduced as special cases by giving 
different values to f, 

In Fig. 166 let AB represent the back of the wall, BD the surface 
of the ground, AD the natural slope, and A Cany line included between 
AB and AD. Also let P' denote the resultant pressure due to the 
wedge BAC, P l the weight of this wedge, OR its reaction against 
the plane AC, the angle between P f and a normal to the back of 
the wall, <w the angle of repose of the earth, a the angle between the 
back of the wall and the horizontal, ft the angle between the surface 
of the ground and the horizontal, and x the angle between AC and 
the horizontal. 

Then in the triangle TOS, by the law of sines, 

, si 

l s 

or, since TOR=x-o> and mS=180-a ?, we have OST=a+-x+a), 

and, consequently, / 

pl = p sin (a; -ft)) 

1 sin (a + f -f- &) x) 

To find an expression for P v let w denote the weight of a unit 
volume of the material, say the weight of one cubic foot. Then for a 
section of unit length in the direction of the wall 

in 
P l = w(area ABC) = -AB - AC sin BAC ; 

z> 

or, if k denotes the height of the wall, AB = -7 > B A C = a - x, 

' ( _ R\ sma 

and AC = AB sm (*-P) . whe nce 
sin (x j3) 

_ wli z sin (a ft) sin (a x) 

2 sin 2 a sin(x ft) 
and, consequently, 

, _ wb? sin (a ft) sin (a x)sin(x ft)) 
~ 2 sin 2 a; siu(x - ft)siu(a ++&)- x) 



256 STRENGTH OF MATERIALS 

The problem now consists in finding the value of the variable angle 
x for which P r is a maximum, which may be expressed symbolically 
by the conditions , p/ 

~ = and 

dx x 

In order to reduce the expression for P' to a form more suitable for 
differentiation, we make use of the following identity. 

cos (a x) cos (a &>) 
cot (a x) cot (a &>) = 

sin (a x) sin (a &>) 

cos (a x) sin (a co) cos (a o>) sin (a x) 

sin (a x) sin (a &>) 
__ sin (x ft>) 

sin (a x) sin (# &>)' 
whence 

sin (x ft>) = sin (a #) sin (a &>) [cot (a x) cot (a o>)]. 

Similarly, 

sin (x P) = sin (a a?) sin (a j3) [cot (a x) cot (# /3)], 
and 

sin (a + &) + f a?) = sin (or #) sin (&> + f) [cot (a x) cot (o> -f ?)]. 
Substituting these values in the expression for P', the latter becomes 
i wh* sin (a CD) cot (a x) cot (a &>) 

~~ 2sin 2 # sin(a)+f) [cot (-#) cot (a /S)][cot(a #)+ cot (&)+?)] 
Now the terms in this expression which contain the variable x are 
all of the same form, namely, cot (a x). This term may therefore 
be replaced by a new variable y t and the remaining terms by letters 

denoting constants. Thus let 

wh 2 sin (a co) 

cot (a x) = y, - = A, 

2 sin 2 a sin (o> + ?) 

cot (a a>)=B, cot (a - /3)= C, cot(&> + ?)= D. 



Equating to zero the first derivative of P' with respect to y y we have 
= ^ (y - C)(y + J)-(y - J3)(y - C)-(y - 

<fy " (y - c-) 2 (2/ + vf 

whence the condition for a maximum is 



= B +(#- C)(B 



FOUNDATIONS AND KETAINING WALLS 257 

Substituting this value of y in the expression for P' t the latter becomes 



or, replacing A, B, C, I) by their values, 

^ sin* (a -co) 
maX 2sin*asin(a + t) , sin (co - /?) sin (co 

" 



which is the general formula for the maximum inclined earth pressure 
against retaining walls. 

The various standard theories as to the maximum eartli pressure 
may now be obtained as special cases of the above general formula 
by making the following assumptions.* 

1. Weyrauch's formula. Assume that the pressure is normal to 
the back of the wall. Then f 0, arid formula (107) becomes 



P' = 



sin 2 (a o>) 



> . a A I sin (ft) /3)sin co\ 2 
2sm 3 o:( 1 -f- A -I 

\ M sm (a p) sin a j 

2. Rankine's formula. Assume that the angle of repose of earth 
on masonry is equal to the angle of repose of earth on earth. Then 
f = co, and formula (107) becomes 

, wh* sin 2 (a co) 1 

max ~2sm 2 tfsin(o; + co) / I sin(o) - ft) sin 2 a> 

3. Poncelet's formula. In Eankine's formula assume that the earth 
surface is horizontal and the back of the wall is vertical. Then 
/3 = and a = 90, and the preceding formula becomes 

f wh 2 cos co 

2(l+V2siii&)) 2 

4. Coulomb's formula. Assume, as in 3, that the earth surface is 
horizontal and the back of the wall vertical, and make the further 

* It is not intended to convey the idea that Weyrauch, Rankine, etc., made these 
assumptions explicitly, but that they lead to formulas identical with theirs. 



258 STRENGTH OF MATERIALS 

assumption that the pressure is normal to the back of the wall. Then 
/3 = 0, a = 90, ?= 0, and formula (107) becomes 

45-i 

5. Rankine's formula for vertical wall Assume that the back of the 
wall is vertical and that the line of action of the resultant earth 
pressure is parallel to the surface of the earth. Then a = 90, 
f 90 + /3 a, and formula (107) becomes 

, _ ivk 2 COS 2 ft) 

^g/i+j^+a^g^jy 

\ M cos 2 /3 i 

6. Maximum normal pressure. Assume that /3 has its maximum 

value, which will be when 
ft = &). Then Wey ranch's 
formula becomes 

f wh 2 sin 2 (a ft>) 
max= 2 sin 3 * 
which is the greatest normal 
thrust that can be caused by 
a sloping bank. 

Problem 293. A wall 20 ft. 
high is inclined at an angle of 85 
to the horizontal and supports a 
backing of clayey gravel the sur- 
face of which makes an angle of 
20 with the horizontal. Compute 
the maximum pressure against the 
back of the wall by Weyrauch's and 
Rankine's formulas, and compare 
the results. 

Problem 294. By the use of 
Poncelet's formula compute the 

-p -g maximum pressure in the preceding 

problem if the back of the wall is 
vertical and the surface of the ground is horizontal. 

Problem 295. What is the greatest normal pressure that can be caused by a 
bank of loose sand against a vertical wall 18 ft. high ? 

172. Stability of retaining walls. The conditions for the stability 
of a retaining wall are the same as those given in Article 155 for the 




FOUNDATIONS AND RETAINING WALLS 259 

stability of abutments, namely, that the wall must be secure against 
sliding on its base and against overturning. 

Let P 2 denote the weight of the wall, P' the resultant earth pres- 
sure, and R the resultant of P 2 and P' (Fig. 167). Then, if R is resolved 
into two components R F and R N , respectively parallel and perpen- 
dicular to the base of the wall, the condition for stability against 
sliding is that R F shall be less than the friction on the base, or, 
symbolically, 

Let g denote the factor of safety. Then this condition may be written 



s r < ks r . 



(108) JB, = r. 

9 

To find the values of R F and R N) let P' and P 2 be resolved into com- 
ponents parallel to R F and R y respectively. Then, in the notation of 
the preceding article, 

R F = P' sin(o; + + f) _ p a sin 6, 
R N = P 2 cos 6 - P' cos(a + 6 + ?). 

Substituting these values of R F and R N in equation (108) and solving 
the resulting expression for g, 



If the base of the wall is horizontal, = and equation (109) becomes 

( no) *[P 2 -P'cos(*+jr)]. 

P'sin(a + f) 

For security against sliding the factor of safety should not be less 
than 3 ; consequently, the criterion for stability against sliding may 

be stated as 

f/>3, 

where the value of g is calculated from equation (109) or (110). 

In applying this criterion it should be noted that the value of f 
must first be assumed (Article 171 ; < ? < a>). 

The following table gives average values of the angle of repose 
and coefficient of friction of masonry on various substances.* 

* See references at the foot of p. 246. 



260 



STRENGTH OF MATERIALS 





MATERIAL 


ANGLE OF 
REPOSE 


COEFFICIENT OF 
FRICTION 


Masonry 

u 


on dry clay 
moist clay 


27 
18 


.510 
325 


u 


wet clay ... 


15 


268 


t( 


dry earth 


30 


577 


u 


clayey gravel 


30 


577 


u 


sand or gravel 


35 


700 


tt 
u 

u 


dry wooden platform 
wet wooden platform . . . 
masonry dry 


31 

37 
31 


.601 
.754 
601 


u 


masonry, damp mortar . 


36 


.726 



In order for a wall to fail by overturning, it must either rotate 
about the outer edge of the base or, in the case of a masonry wall, 
open at one of the joints. The cause of failure in both cases is the 
same, namely, that the stress on the base or joint is partly tensile. 
Consequently, the criterion for stability against overturning is that 
the resultant R must strike within the middle third of the base or 
joint, as the case may be (cf. Articles 62, 148, 2, and 155). 

This criterion can best be applied graphically. Thus having assumed 
a value for the angle f, the resultant earth pressure P' is calculated 
from the formula in Article 171, corresponding to this assumption 
of f, and combined with the weight of the wall into a single result- 
ant R. If this resultant does not strike within the middle third of 
the base, or within the middle third of all the joints in the case 
of a masonry wall, the design must be altered until the criterion is 
satisfied. 

173. Thickness of retaining walls. In designing a retaining wall 
economy of material is secured by making the base of such thickness 
that the resultant R, obtained by combining the weight of the wall 
P 2 with the maximum earth pressure P 1 ', shall fall at the outer edge 
of the middle third. However, theoretical formulas for determining 
the least thickness consistent with this condition are too complicated 
to be of practical value, and for this reason the design is usually based 
on an empirical formula. 

In railroad practice Trautwine recommends that for vertical walls 
of rectangular cross section, supporting loose sand, gravel, or earth 



FOUNDATIONS AND RETAINING WALLS 261 

level with the top, the thickness 1} of the base of the wall in terms 
of its total height li should be as follows : * 

For wall of cut stone or. large ranged nibble f in mortar, 

1> = .35 h. 
For wall of good common 8cabblcd mortar rubble, or brick, 

1> = .40 h. 
for wall of well scabbled dry rubble, 

1> = .50 h. 

These empirical rules may be regarded as representative of the best 
American practice, and may be used to give a first approximation in 
making a tentative design. 

By inclining the wall backward the angle between the earth thrust 
P' and the wall is decreased, and consequently the resultant li is 
made to approach more nearly the center of the base. This allows 
the thickness of the base to be decreased and thus lessens the 
amount of material in the wall, although it slightly increases its 
depth. However, there is a restriction upon the amount of inclina- 
tion which is permissible, for the inclination also has the effect of 
increasing the tendency to slide on the base or joints. In practice 
these considerations are balanced by inclining the back of the wall 
at a small angle, say 5 or 10, to the vertical (i.e. a = 80 or 85), 
and at the same time cutting the footing into steps perpendicular 
to the line of action of the resultant R, thus securing economy of 
material without sacrificing stability. 

The thickness of the top of the wall is determined by the necessity 
of providing for the lateral pressure of the earth, due to the action of 
frost. Since the action of frost is greatest near the top of the wall 
where the material is most exposed, it is likely to push the top over 
if the wall is made only thick enough to resist the pressure due 
to the weight of the earth. This consideration, therefore, limits the 
least thickness of the wall at the top to about two feet for masonry, 
or somewhat less than this amount for concrete, since the latter has 
no joints and therefore offers a greater moment of resistance. 

* Engineer's Pocket-Book, 1902, p. 603. 

t Masonry composed of rough, undressed stones is called rubble ; scabbled rubble has 
the roughest irregularities knocked off with a hammer. 



262 STRENGTH OF MATERIALS 

From the above, it follows that for an economical design the cross 
section of a wall should be trapezoidal, the thickness of the base 
being determined by the consideration of stability against overturn- 
ing, and the thickness of the top by the maximum action of frost. 

The inclination of either face of a wall to the horizontal is usually 
expressed by giving the ratio of the horizontal projection of this face 
to its vertical projection. This ratio is called the batter, and is given 
in inches of horizontal projection per foot of height. For example, if 
a wall makes an angle of 80^- with the horizontal, it is said to be 
" battered 2 to 1," since the ratio of its horizontal projection to its 
vertical projection is equal to cot a, and in the present case 

cot a = cot 80^ = .1673 = ^, approximately. 

Problem 296. Design a concrete retaining wall to support a bank of loose earth 
25 ft. high, the back of the wall to be inclined backward at a batter of H to 1. 



PART II 
PHYSICAL PROPERTIES OF MATERIALS 



263 



PART II 

PHYSICAL PROPERTIES OF MATERIALS 
CHAPTER XII 

IRON AND STEEL 

174. Introductory. A study of the properties of materials used in 
engineering construction involves a study of the machines used for 
making the tests and the method of conducting these tests. From 
the time of Galileo, in 1600 A.D., tests have been made to determine 
the strength of materials, but only during the past fifty years has any 
very great advance been made. The rapid development of the past 
half century has been due to the notable increase in the construction 
of large buildings, bridges, etc. ; for where engineers were formerly 
content to use material without being tested, the importance of modern 
constructions demands that the physical properties of the materials 
used shall be determined for each large contract. 

The early testing machine consisted of little more than an ordinary 
scalebeam with the test piece attached to one end and the load 
applied at the other. These were used for making tension tests, and 
machines equally as simple were used for compression and flexure 
tests. Fig. 168 shows a type of these machines which was used by 
Kirkaldy about 1860. The specimen to be tested was held in the 
jaws g while the lever F was in the position of the dotted lines 
(Fig. 168). A load N was then applied to the end of the lever and 
gradually increased until the specimen was ruptured. 

Testing machines have been much improved during the past twenty 
or thirty years in the United States by Eiehle Bros, and Olsen & Co., 

265 



266 



STRENGTH OF MATERIALS 



both of Philadelphia, Pennsylvania. The machines as now con- 
structed for ordinary testing purposes consist of a platform scales 
with the usual means of measuring loads, and a screw press operated 
by an outside source for applying the loads. Fig. 169 is a machine 
of 100,000 Ib. capacity, built by Olsen & Co., and may be taken 
as a type. The four upright pieces A with the base B upon which 
they rest form the platform of the scales. This platform rests upon 
knife-edges C attached to a system of levers D which terminate 
finally in a graduated lever E (the scalebeam) provided with a 
movable poise. Each lever is supported by knife-edges resting upon 
hardened steel plates. The screw press in this case is seen in the 




FIG. 168 

four screws F with their movable crosshead G. The upper cross- 
head H is attached to the four upright pieces and is a part of the 
scale platform. 

175. Tension tests. If a piece is to be tested in tension, one end 
is attached to the upper crosshead and the other end to the lower. 
The turning of the screws, due to the driving mechanism on the 
other side of the machine, causes the lower crosshead to move down- 
ward, thus bringing pressure to bear on the upper crosshead. From 
here it is transmitted to the base and thence to the levers, and is 
measured by movement of the poise on the graduated scalebeam. 
Machines of 20,000 Ib., 30,000 Ib., 50,000 Ib., 100,000 Ib., 200,000 Ib., 
and 300,000 Ib. capacity are manufactured, as well as a great many 
machines for making special tension tests. In the larger testing 
machines the upper head is usually adjustable so as to accommodate 
specimens of various lengths, but in the smaller machines the upper 
head is fixed. 




267 



IRON AND STEEL 269 

The tensile strength in pounds per square inch is computed by 
dividing the load read from the scalebeam by the area of cross section 
of the test specimen (see Article 20). Expressed as a formula, 

load from scalebeam 
Tensile strength in lb./in. 2 = 



area of cross section 

176. Compression tests. To make compression tests the piece is 
placed on a small block resting on the platform, and the lower cross- 
head, provided with a similar block, is brought down upon it. The 
further lowering of the crosshead compresses the specimen. The pres- 
sure comes on the platform through the crossbeam that rests upon it, 
and is transmitted to the scalebeam, where it is measured. 

The compressive strength in lb./in. 2 is computed by dividing the load 
in pounds as read on the scalebeam by the area of cross section of 
the test specimen, as in finding the tensile strength. 

177. Flexure tests. Beams are tested in flexure by mounting the 
specimen on a crossbeam provided with knife-edges and applying the 
load from above by means of a knife-edge attached to the under side 
of the moving head. The beam is tested by lowering the moving 
head as in the compression tests. 

The fiber stress in the outer fiber of the beam is computed in this 
case from the formula (see Article 52), 

Pie 



where e is the distance from the neutral axis to the outer fiber, / is 
the moment of inertia with reference to the neutral axis, P is the 
load in pounds as read from the scalebeam, I is the length of the span 
in inches, and p is the fiber stress in lb./in. 2 

The maximum deflection for the concentrated central load is com- 
puted by the formula (see Article 67), 



where D is the deflection at the center, E the modulus of elas- 
ticity (see Article 8), and P, I, and I have the same meaning as 
above. 




270 STRENGTH OF MATERIALS 

In case the beam is loaded at the third points, uniformly, eccen- 
trically, or otherwise, the corresponding expressions are used for fiber 
stress and deflection (see Articles 52, 67). 

178. Method of holding tension specimens. To make a tension 
test of a material a special test piece is usually provided. This test 
piece has the same composition as the rest of the material, but has a 
special form, being larger at the ends than in the central portion (see 
Article 20). Fig. 170 illustrates a test piece made from a carbon 
steel bar turned down in the central portion.* The machines are 

provided with serrated wedges 
for holding the large ends of the 
test piece, and as the load is ap- 
plied these serrations sink into 
the specimen, thus holding it 
firmly. 

The behavior of the specimen 
in tension is studied by noting 
FlG> m the behavior of the reduced 

portion, which should be far enough from the ends so that the local 
stress caused by the wedges will have no effect upon it. 

Flat pieces, such as pieces of boiler plate, are left as they come 
from the rolls on two sides, and the edges are machined to get the 
reduced cross section, as shown in Fig. 171. The lower specimen, 
of cast iron, is made with rounded corners to eliminate shrinkage 
stresses. Eolled material is often tested without being turned down. 
Special holders and clamps are usually provided for holding tension 
specimens of timber. 

179. Behavior of iron and steel in tension. Wrought iron and mild 
steel when tested in tension conform to Hooke's law up to the elastic 
limit, a point which is usually well defined in these materials. They 
then suffer a rapid yielding, with little increase of load, reaching a 
point where the piece elongates very much for no increase of load. 
This point is known as the yield point. It is indicated by the scaling 
of the oxide from the specimen that has not been machined, and by 
the dropping of the beam of the testing machine, if it has been kept 
balanced up to this point. Beyond this point stress increases much 

* Dimensions for standard test specimens of different materials are given in Article 203. 



IRON AND STEEL 271 

more slowly than deformation, until finally rupture is about to 
occur, at which point the load attains its maximum value, called 
the ultimate load. If the stress be continued, the piece begins to 
neck and breaks at a load somewhat less than the maximum (see 
Article 7). This necking is due to the fact that the metal under 
great strain becomes plastic and flows. Brittle materials, such as 
cast iron and hard steel, show very little, if any, necking. In com- 
puting the fiber stress at the maximum load the original cross section 
is used. 

In commercial tests the load at the yield point (commercial elastic 
limit) and the maximum load are noted ; also the percentage of 
elongation and the percentage of reduction of cross section. The per- 
centage of elongation is the increase in length divided by the original 
length multiplied by 100. This percentage varies with the original 
length taken (see Article 20), and therefore is usually computed for an 
original length of eight inches. The percentage of reduction of cross 
section is the decrease in area of the cross section divided by the 
original area of the cross section multiplied by 100. In some com- 
mercial laboratories provision is made for making as many as sixty 
tests per hour on one machine. 

180. Effect of overstrain on wrought iron and mild steel. If 
wrought iron and mild steel are strained just beyond the elastic limit 
in tension or compression, then released and tested again in the same 
direction, it has been found that this second test shows that the 
elastic limit is higher than at first, and almost as high as the load 
in the first test. Eepeated overstrain of this kind, with subsequent 
annealing, makes it possible to raise the elastic limit considerably 
above what it was originally. When further strained the metal 
loses its elasticity and takes on a permanent set ; that is to say, it 
does not return to its original length when the stress is removed. 
The elastic properties, however, can be restored by annealing (see 
Article 18). Overstrain in either tension or compression destroys 
almost entirely the elasticity of the material for strain of the opposite 
kind ; for instance, a piece of mild steel overstrained in tension has its 
elastic properties in compression almost entirely destroyed, and vice 
versa. Overstraining in torsion produces much the same effect as 
overstraining in tension or compression. 



272 STRENGTH OF MATERIALS 

181. Relative strength of large and small test pieces. It has been 
found by Tetinajer * and others that the values obtained in testing 
small test pieces taken from different parts of a steel girder or I-beam 
are higher than those obtained in testing the girder itself. The aver- 
age of a series of tests of small test pieces gave an elastic limit of 
49,000 lb./in. 2 and a maximum strength of 62,000 lb./in. 2 Tests on 
the complete girders themselves gave an elastic limit of 33,500 lb./in. 2 
and a maximum strength of 54,500 lb./in. 2 The same has been found 
true for the elastic limit of wrought-iron girders, but in this case the 
maximum strength is greater in the girder than in the small test piece. 

182. Strength of iron and steel at high temperatures. From a 
series of tests made at Cornell University,! it was found that wrought 
iron having a tensile strength of 30,000 lb./in. 2 at ordinary tem- 
peratures increased in strength with increase of temperature up to 
475 F., and then decreased as the temperature was further raised. 
Machinery steel of 60,000 lb./in. 2 maximum strength gave at 475 F. 
a maximum strength of 111,500 lb./iu. 2 Tool steel having a strength 
of 114,000 lb./in. 2 at ordinary temperatures gave 145,000 lb./in. 2 
maximum strength at 350 F. 

Professor C. Bach also reports an elaborate series of tests on the 
strength of steel at high temperatures. $ At ordinary temperatures 
one bar had a maximum strength of 54,000 lb./in. 2 , an elongation in 
8 in. of 26.3 per cent, and a contraction of area of 46.9 per cent. 
Up to a temperature of 572 F. the strength increased by about 
7000 lb./in. 2 , and from this point fell, approximately in proportion to 
the temperature, to 26,200 lb./in. 2 at 1022 F. The ultimate elonga- 
tion decreased to 7.7 per cent at 392 F., and then increased to 39.5 
per cent at 1022 F. The contraction of area fell until 392 F. was 
reached, and did not rise until about 572 F. 

While the tensile strength is increased for a moderately high tem- 
perature, the elastic limit is lowered in proportion to the increase of 
temperature, being diminished about 4 per cent for each increase of 
100 F. 

183. Character and appearance of the fracture. The kind and 
quality of the metal are usually indicated by the character of the 

* Communications, Vol. IV. f Journal Western Society of Engineers, Vol. I. 

J Journal Franklin Institute, December, 1904. 



IKON AND STEEL 273 

fractured portion of the test piece. Two points are to be noted in 
this connection : the geometrical form and the appearance of the fractured 
material. Under the first we may have, as in tensile tests of hard 
steel, a straight fracture where the material breaks squarely off in a 
plane at right angles to the axis of the test piece ; or, as in tensile tests 
of mild steel and high-grade wrought iron, a fracture which is cup- 
shaped, half-cup, etc. The appearance of the material for the cup-shaped 
fracture may be described as dull granular in the bottom of the cup 
and silky around the edge ; or, in the case of wrought iron, as fibrous 
in the bottom of the cup and silky around the edge. A cast-iron 
fracture appears crystalline, the crystals being fine, coarse, or medium. 

In reporting a test the character and appearance of the fracture 
should always be given. It should also be noted whether or not any 
longitudinal seams occur, or whether the fracture shows the material 
to be homogeneous and free from blowholes and foreign matter. If 
the specimen has not been properly placed in the machine, so that 
there is a bending moment, the fracture will indicate this. The axis 
of the test piece should always coincide with the axis of the machine. 

184. Measurement of extension, compression, and deflection. The 
extension in a tension specimen of iron or steel up to the elastic 
limit is so slight that very accurate measurements must be made to 
determine the elongations. Instruments for making such measure- 
ments are known as extensometers, and are usually made to read to 
.0001 of an inch. Fig. 173 shows a type of such instrument known 
as the Yale-EieJile extensometer. The method of using the instrument 
is to mark off an 8-in. gauge length on the test piece and fasten 
the extensometer to it by inserting the screws in the extreme punch 
marks of the gauge length. The backpiece is then removed and a 
battery with a bell in circuit is attached ; the instrument is then 
ready for use. As the piece elongates the elongations are measured 
by turning the micrometer screw until it touches the armature, when 
the circuit is closed and the bell rings. 

The instrument is used only a little past the elastic limit (the limit 
of proportionality of stress to deformation), and about twenty elonga- 
tions for corresponding loads are taken below the elastic limit. The 
instrument is then removed and the test continued to failure, the 
maximum load being noted. From the data obtained in making 



274 STRENGTH OF MATERIALS 

the test, the strain diagram is drawn by using unit loads as ordinates 
and relative elongations as abscissas. From this curve the elastic limit, 
modulus of elasticity (Young's modulus), and modulus of elastic 
resilience may be determined. 

The elastic limit is found by noting the point on the strain diagram 
where it ceases to be a straight line. 

The modulus of elasticity is determined by dividing the stress by 
the deformation for any stress below the elastic limit. 

The modulus of elastic resilience is denned as the amount of work re- 
quired to deform a cubic inch of the material to its elastic limit. It is 
therefore represented by the area under the strain curve up to the elastic 
limit, or, expressed as a formula, 

(stress at elastic limit) 2 

Mod. elastic resilience = 

2 modulus of elasticity 

If in plotting the strain diagram the ordinates represent the stress 
expressed in lb./in. 2 and the abscissas represent the correspond- 
ing unit elongations, the area under the curve up to the 
elastic limit multiplied by the scale value in inch-pounds of 

each unit area gives 
the modulus of elas- 
tic resilience in inch- 
pounds. 

The modulus of total 
FIG. 172 

resilience is denned as 

the amount of work required to deform a cubic inch of the material 
to rupture. It is therefore represented by the area under the whole 
curve multiplied by the scale value of a unit area, that is, the number 
of inch-pounds per unit area. 

In case the stresses are plotted in pounds and the corresponding 
deformations in inches, the above method gives the work done on the 
whole volume of the specimen included in the gauge length. To 
obtain the modulus for such cases it is necessary, in addition to the 
above, to divide by the volume of that portion of the specimen over 
which the deformations were measured. 

Compression is measured by means of a compressometer, by methods 
similar to those used in making tension tests. The strain diagram in 
this case is a stress-compression curve. 




IKON AND STEEL 



275 



For measuring deflections in transverse tests various methods are 
used. A simple instrument for this purpose is shown in Fig. 172. 
This instrument is placed under the beam and the deflections meas- 
ured to .001 of an inch. The strain diagram for flexure is thus a 
load-deflection curve. 

Problem 297. A rod of nickel steel .854 in. in diameter, and with a gauged 
length of 8 in., when tested in tension gave the data tabulated below. From this 
data draw the strain diagram and locate the elastic limit ; also compute the mod- 
ulus of elasticity and the modulus of elastic resilience. 



LOAD 
lb./in. 2 


ELONGATION 
inches pei' inch 


LOAD 
lb./in.2 


ELONGATION 
inches per inch 


4,000 


.00018 


36,000 


.00096 


8,000 


.00032 


40,000 


.00103 


12,000 


.00040 


44,000 


.00115 


10,000 


.00050 


48,000 


.00125 


20,000 


.00060 


52,000 


.00165 


24,000 


.00065 


56,000 


.00470 


28,000 


.00075 


91,400 


Maximum load 


32,000 


.00083 







185. Torsion tests. The determination of the resistance of a material 
to shear or torsion is usually made by means of a machine designed 
to read twisting moment in inch-pounds on the scalebeam. The 
Biehle machine shown in Fig. 174 may be taken as a type of torsion 
machines. 

In making the test one end of the specimen is attached to the 
twisting head of the machine and the other end to the stationary 
head, which is connected by a system of levers to a scalebeam read- 
ing inch-pounds of moment. The machine shown in Fig. 174 has 
the stationary head suspended by stirrups, thus leaving it free to move 
slightly when the specimen shortens in twisting. The older types of 
torsion machine are not made to accommodate themselves in this way 
to the shortening of the test piece. 

The angular distortion of the test bar is measured by an instrument 
called a troptometer. This consists of two arms attached to the bar at 
the extreme points of the part that is being tested. One of these 
arms carries a scale bent into the arc of a circle of which the arm is 



276 



STEENGTH OF MATERIALS 



the radius, and having its plane at right angles to the axis of the 
bar; the other arm carries a pointer so arranged as to move over 
the scale when the bar is twisted. The arc of the scale is called the 
troptometer arc and the arm supporting it the troptometer arm. The 
angular distortion at the center of the bar for the given gauge length 
is then obtained by dividing the reading on the troptometer arc by 
the length of the troptometer arm plus the radius of the specimen, 
or, expressed as a formula, 

reading" on troptometer arc 



Ang-le 8 (in radians) = 



troptometer arm + radius of specimen 



where is the angle of twist (see Article 96). 

Problem 298. A steel rod with a gauged length of 10 in. and .85 in. in diam- 
eter, when tested in torsion, gave the data tabulated below. Draw the strain 
diagram, plotting the stress in lb./in. 2 on the outer fiber as ordinates, and the cor- 
responding angle of twist 6 as abscissas. Also locate the elastic limit, compute the 
modulus of elasticity of shear, and the modulus of elastic resilience. Length of 
troptometer arm, 12 inches. 

TORSION TEST OF STEEL 



MOMENT 
in. Ib. 


TKOPTOMETER ARC 
in. 


MOMENT 
in. Ib. 


TROPTOMETER ARC 
in. 








1750 


.33 


250 


.05 


2000 


.38 


500 


.10 


2250 


.43 


750 


.15 


2500 


.47 


1000 


.20 


2750 


.53 


1250 


.24 


3000 


.57 


1500 


.29 







186. Form of torsion test specimen. Specimens for torsion tests 
are made cylindrical, and usually long enough to get a gauged length 
of 10 in. The cylindrical form has been adopted because its cross 
sections remain plane during torsion, whereas in other forms a cross 
section which is plane before torsion is deformed into a warped 
surface by the strain, and therefore does not give a simple shearing 
stress (see Article 102). The torsion test is used to determine the 
shearing strength of materials, that is, the resistance offered by tli3 
material to one cross section slipping over another (see Article 68). 



IKON AND STEEL 277 

When torsion tests are made, the moment in in. Ib. is read from 
the machine, and the shearing stress in the outer fiber in lb./in. 2 is 
computed from the formula, 



p 



where Pa is the twisting moment read from the machine, r the 
radius of the test piece, and I p the polar moment of inertia of the 
cross section.* 

The modulus of elasticity in shear is computed from the relation, 




where Pa and I p are defined as above, I is the gauged length in inches, 
and is the angle of twist in radians. 

The test piece is held in position by a set of adjustable jaws similar 
to those used in ordinary pipe wrenches. The gauged length should 
be taken far enough from the ends so that the local stress due to the 
jaws may not influence the results. 

187. Torsion as a test of shear. Although the torsion test is used 
to determine the shearing strength of materials, it is not an accurate 
test, since the shearing stress is a maximum on the outer elements, 
and zero at the center. For this reason the inner material tends to 
reenforce the outer, thus giving a higher shearing strength than 
would otherwise be obtained. A more perfect torsion test would be 
one made upon a hollow tube of the material, for in this case the 
inner reenforcing core would not be present. However, the difficulty 
of obtaining suitable hollow tubes for test pieces makes their use 
impracticable for ordinary tests. 

A further objection to the torsion test as a test of shearing strength 
lies in the fact that there is considerable tension in the outer ele- 
ments of the test piece during the test. Any element of the cylin- 
drical test piece which is a straight line before the strain becomes a 
helix during the test. Since the length of the helix is greater than 
that of the original element, a tensile stress is thus produced in the 
outer fibers. In fact, in testing wrought iron in torsion the outer 
fibers often fail in tension along the helix. The slight shortening 

TTr 4 
* For a cylinder, I p = 



278 STRENGTH OF MATERIALS 

of the whole specimen, due to the twisting, is corrected in part by 
the swinging head of the machine shown in Fig. 174. 

188. Shearing tests. To determine the shearing strength of timber 
along the grain and the resistance of iron and steel to the pulling out 
of rivets, many special tests are used. By means of a special piece of 
apparatus, the force required to push off, along the grain, a projecting 
piece from a test piece of timber is easily measured on the ordi- 
nary tension-compression machine. The intensity of shearing stress is 
computed by dividing the load by the area of the block pushed off. 

Tests are also made on wrought-iron plates to determine the force 
required to pull out a rivet through the metal, both in the direction 
of the fiber and perpendicular to it. A series of such tests may be 
found in the Watertown Arsenal Report for 1882. Many tests have 
also been made to determine the shearing strength of rivets. 

189. Impact tests. In actual service many materials are subjected 
to shock or impact (see Article 74). This is especially true of all 
railway structural material, such as rails, axles, springs, couplers, 
bolsters, wheels, etc., which must be designed to withstand consid- 
erable shock. Two special machines have been designed to test 
materials in impact. The first, called the drop testing machine, is 
operated by allowing a given weight (hammer) to drop a given dis- 
tance upon a test piece mounted on an anvil under the hammer. 
The other form of machine is operated by allowing a heavy pendu- 
lum to strike the specimen when placed in the center of its swing. 
In either case the amount of the energy of the blow absorbed by the 
specimen is desired. 

The results obtained from impact tests can only be comparative in 
any case, since a part of the energy of the blow must be absorbed 
by the parts of the machine itself. This is seen in the drop testing 
machine in the absorption of energy by the anvil and hammer. 

Since the results of such tests cannot be absolute, it is highly 
necessary that they should be standardized by making tests on the same 
anvil with the same hammer. The Master Car Builders Association 
has taken a step toward such standardization by building an impact 
testing machine for testing materials used by them. This machine 
has been established at Purdue University. Its maximum blow is 
given by a hammer having a weight of 1640 lb., and dropping 50 ft. 



IRON AND STEEL 279 

The use of this machine should do much to standardize specifications 
for railway material.* 

Tests in impact compression, impact tension, and impact flexure 
are also made, but on account of the uncertainty as to the amount of 
energy absorbed by the test specimen many engineers do not favor 
such tests. Many of these objections, however, might be removed by 
proper standardization. 

Some recent investigations seem to indicate that the impact test 
shows very little that cannot be determined by static tests. 

190. Cold bending tests. Cold bending tests are tests of the duc- 
tility of metals, and are designed to show the effect on the metal of 
being bent in various ways while cold. Such material as rivet steel 
and Bessemer steel bridge pieces are bent double over a pin of speci- 
fied radius, and the result noted. In making these tests the angle at 
which the first crack occurs and the angle at w r hich rupture occurs 
are read. 

Few machines for making cold bending tests have been made. 
The tests are usually made by bending the specimen over the edge 
of a vise, or some such simple device, according to specifications. The 
tests have never been standardized, but their importance is obvious, 
since the conditions of actual service are thus applied to the specimen. 

191. Cast iron. Pig iron is a combination of iron with small 
percentages of carbon, silicon, sulphur, phosphorus, and manganese, 
obtained from the blast furnace. The carbon probably comes from 
the fuel used in reducing the ore ; the other impurities come either 
from the ore or from the flux. The product is graded, according to 
chemical composition, into forge pig and foundry pig. Foundry pig 
is remelted in a cupola furnace and made into castings of various 
kinds ; forge pig is used in making wrought iron. 

Cast iron is a very brittle material, weak in tension and strong in 
compression. Its great usefulness in engineering structures comes 
from the fact that it may be readily molded into any desired form ; 
it is, however, being replaced by the various steel products. The 
carbon, silicon, and other impurities contained in the iron affect its 
physical properties. 

* For a description of this machine see the report by Professor W. F. M. Goss, Proc. 
Amer. Soc. for Testing Materials, 1903. 



280 



STRENGTH OF MATERIALS 



COMPOSITION AND TENSILE STRENGTH OF CAST IRON 
WATERTOWN ARSENAL REPORT, 1895 



CHEMICAL COMPOSITION 


TENSILE 
STRENGTH 


CARBON 












Graphitic 


Combined 


Manganese 


Silicon 


Sulphur 


Phosphorus 


lb./in. 2 


2.917 


.570 


.464 


1.457 


.122 


.539 


28,980 


2.367 


.529 


.458 


1.022 


.120 


.379 


28,620 


2.017 


.710 


.438 


1.193 


.125 


.350 


27,240 


2.691 


.394 


.439 


1.494 


.104 


.538 


24,970 


2.786 


.794 


.446 


1.372 


.100 


.521 


24,880 


2.765 


.589 


.437 


1.457 


.093 


.537 


29,720 


2.140 


1.132 


.445 


0.705 


.104 


.504 


32,020 


2.372 


1.006 


.436 


0.921 


.105 


.473 


33,500 


2.356 


0.952 


.432 


1.071 


.100 


.531 


33,400 


2.263 


1.009 


.451 


0.846 


.108 


.505 


32,010 


2.247 


0.867 


.441 


0.977 


.110 


.472 


32,990 


2.160 


1.068 


.435 


0.864 


.095 


.493 


32,280 


2.208 


0.982 


.430 


0.874 


.096 


.498 


32,200 


2.266 


1.180 


.426 


0.893 


.102 


.504 


30,400 


2.225 


1.074 


.451 


0.902 


.095 


.473 


32,510 



Carbon occurs as combined carbon or as graphitic carbon. Com- 
bined carbon makes the metal hard, brittle, white, weak in tension, and 
strong in compression, whereas graphitic carbon makes the iron soft, 
gray, and weak in both tension and compression. Graphitic carbon 
occurs in the metal as a foreign substance, which probably accounts 
for its weakening effect. Silicon in cast iron up to 0.5 per cent in- 
creases its compressive strength. The tensile strength is increased up 
to 2 per cent. Manganese as it usually occurs is not injurious below 
1 per cent. When more is present the shrinkage, hardness, and brittle- 
ness are rapidly increased. Phosphorus makes the iron weaker and 
less stiff, becoming a serious impurity when it occurs in quantities 
above 1.5 per cent. Sulphur causes whiteness, brittleness, hardness, 
and greater shrinkage, and is, in general, a very objectionable impurity. 

Cast iron has an average tensile strength of 22,500 lb./in. 2 , the 
range being from 13,000 lb./in. 2 to 35,000 lb./in. 2 Its compressive 



IKON AND STEEL 



281 



36 



r, 



30 



24 



21 



18 



15 



12 



STRAIN DIAGRAM 
TENSION TEST OF CAST IRON 



.001 .002 .003 .004 .005 .006 .007 .008 

FIG. 175 



282 STRENGTH OF MATERIALS 

strength varies from 50,000 lb./in. 2 to 150,000 lb./in. 2 , a good average 
being about 95,000 lb./in. 2 

The metal is so imperfectly elastic that Hooke's law does not 
strictly hold for any range of stress, however small. The modulus of 
elasticity in tension varies from 15,000,000 to 20,000,000 lb./in. 2 , and 
in shear from 5,000,000 to 7,000,000 lb./in. 2 On page 280 is given 
a table of the tensile strength of various samples of cast iron of 
different chemical compositions. 

192. Strain diagram for cast iron. The strain diagram of cast 
iron in tension, shown in Fig. 175, illustrates clearly the fact that 
the metal is very imperfectly elastic. No part of the diagram is a 
straight line, and no elastic limit is shown by the curve. The maxi- 
mum load in this case was 34,750 lb./in. 2 The curve was drawn 
from data given in the Watertown Arsenal Report, 1895. From the 
results of four hundred and fifty tests of cast iron in tension, com- 
pression, and cross-bending, Kirkaldy found the average compressive 
strength to be 121,000 lb./in. 2 , the tensile strength 25,000 lb./in. 2 , 
and the cross-bending modulus (see Article 65) 38,000 lb./in. 2 

Fig. 176 shows a strain diagram of cast iron in compression. Like 
the tension diagram, this shows no well-defined elastic limit and no 
constant modulus of elasticity. The maximum compressive strength 
in this case was 50,000 lb./in. 2 

When tested in compression as a short block, cast iron has a 
characteristic fracture, shearing along a plane making an angle of 
about 30 with the vertical. This differs by 15 from the theoretical 
angle (45) of maximum stress for such cases. 

193. Cast iron in flexure. The most extended series of tests ever 
made on cast iron in flexure was made by J. W. Keep on bars J- in. 
square and 12 in. long. From these tests, the average strength was 
found to be 450 lb., giving a modulus of rupture of 64,800 lb./in.' 2 
A good average for the modulus of rupture for ordinary commercial 
cast iron would be between 36,000 lb./in. 2 and 42,000 lb./in. 2 

194. Cast iron in shear. The strength of cast iron in shear varies 
from 13,000 lb./in. 2 to 25,000 lb./in. 2 Tests are made in the ordinary 
torsion machine. The fracture in this case is the characteristic frac- 
ture of brittle materials in torsion ; that is, instead of shearing off in a 
plane at right angles to the axis of the test piece, as is the case with 



IRON AND STEEL 



283 



STRAIN DIAGRAM 
COMPRESSION TEST OF CAST IRON 




,007 



284 



STRENGTH OF MATERIALS 



ductile materials, the fracture extends down one side for some dis- 
tance. The material fails by the outer fiber failing first in tension. 
A similar fracture can be seen by twisting a stick of chalk or other 
brittle material with the fingers until fracture occurs. 

195. Cast-iron columns. Some tests have been made upon full-sized 
cast-iron columns both at the Watertown Arsenal and by the Phoenix 
Iron Company of Phoenixville, Pennsylvania. The results of these 
tests show that the total strength of these columns is much less 
than the compressive strength of the metal would lead one to expect. 
This was probably due to the presence of blowholes or other imper- 
fections in the column, such as are likely to occur when large pieces 
are cast. The ultimate strength of the Watertown columns varied 
from 21,0001b./in. 2 to 40,000 lb./in. 2 

The following table gives the result of a compression test of a cast- 
iron column made by the Watertown Arsenal, the ultimate strength 
in this case being 33,340 lb./in. 2 



COMPRESSION TEST OF CAST-IRON COLUMN 

Gauge length, 100 in. Sectional area, 17 in. 2 
WATERTOWN ARSENAL REPORT, 1893 



LOAD 


COMPRES- 


DEFLEC- 
TION AT 


LOAD 


COMPRES- 


DEFLEC- 
TION AT 




lb./in.2 


SION 
in. 


MIDDLE 
in. 


Ib./in.s 


SION 
in. 


MIDDLE 
in. 













18,000 


.1390 


.05 




500 








20,000 


.1597 


.06 




1,000 


.0032 





22,000 


.1816 


.08 




2,000 


.0093 





24,000 


.2080 


.10 




4,000 


.0225 





26,000 


.2430 


.12 




6,000 


. .0373 


.01 


28,000 




.17 




8,000 


.0530 


.02 


30,000 




.24 




10,000 


.0688 


.02 


32,000 


. 


.40 




12,000 


.0853 


.02 


33,000 




.66 




14.000 


.1023 


.03 


33,340 




1.10 


Ultimate strength 


10,000 


.1204 


.04 

























Problem 299. The data in the preceding table were obtained from a round, 
hollow, cast-iron column 120 in. in length, 3.05 in. in external diameter, and 
1.97 in. in internal diameter. Draw the load-compression and load-deflection 



IKON AND STEEL 



285 



curves for this case, and determine whether or not an elastic limit is indicated. 
Also compute the strength of the column by Rankine's formula and Johnson's 
straight-line formula, and compare the results with those obtained from the test. 

196. Malleable castings. The castings with combined carbon are 
hard and brittle. These are heated with some oxide, so that the 
carbon near the surface is burned out, leaving the outer surface 
tough and strong, like wrought iron. The interior of the casting 
is somewhat annealed, but the finished product consists of a hard 
interior portion with a ductile outer portion. This structure insures 
strength both statically and as regards impact. 

197. Specifications for cast iron.* The following specifications 
are for special hard cast iron (close-grained). They are taken from 
the J. I. Case Threshing Machine Company's specifications, and may 
be considered as typical. 

CHEMICAL COMPOSITION 

Silicon must be between 1.20 and 1.60 per cent. (Below 1.20 the 
metal will be too hard to machine ; above 1.60 it is likely to be 
porous unless much scrap is used.) 

Sulphur must not exceed 0.095 per cent, and any casting showing 
on analysis 0.115 per cent or more of sulphur will cause the rejec- 
tion of the entire mixture. (Above 0.115 per cent sulphur produces 
much shrinkage, shortness, and " brittle hard " iron.) 

Phosphorus should be kept below .70 per cent unless specified for 
special thin castings. (High phosphorus gives castings brittle under 
impact.) 

Manganese should not be above .70 per cent except in special chilled 
work. 

PHYSICAL TESTS 

Transverse breaking strength. The test bars should be 1 in. square 
and 13^ in. long, and should be tested with a load of 2400 Ib. applied 
at the center of a 12-in. span. 

* These specifications, as well as all others quoted, are given so that the student may 
get an idea of the composition and properties required of commerial cast iron or other 
material. Specifications issued by different companies vary, and those issued by the 
game company are frequently changed on account of the requirements of service. 



286 



STEENGTH OF MATERIALS 



Deflection should not be less than 0.08 in. 

Tensile strength must not be less than 22,000 lb./in. 2 

The following specifications for cast iron are suggested by J. W. 
Keep as being representative of modern practice.* 

Transverse test bars were cast 1 in. square and 12 in. long, and 
were tested with a central load. Tensile test bars were cast 1.13 in. 
in diameter and were tested as cast. 



CHARACTER OF CASTING 


SILICON 
RANGE 


SUL- 
PHUR 

BELOW 


PHOS- 
PHORUS 

BELOW 


MAN- 
GANESE 

BELOW 


TRANS- 
VERSE 
STRENGTH 
Ib. 


TENSILE 
STRENGTH 
lb./in .2 


Furnace | He * Vy ' ' 

1 ATfxliiiin 


1.20-1.50 
1.50-2.00 

1.20-1.50 
1.50-2.00 


.085 
.085 

.090 
.080 


.24 

.68 

.60 
.60 


.37 
.04 

.80 
.80 


3900 
3200 

2600 
2400 


38,000 
31,000 

25,000 
23,000 




. f Heavy . 
S P 6Cial | Medium 


Cupola < 


(-Heavy . 
General-j Medium 
L -Light . 


1.20-1.75 
1.40-2.00 
2.20-2.80 


.090 
.085 
.085 


.70 
.70 
.70 


.70 
.70 
.70 


2400 
2200 
2000 


22,000 
20,000 
18,000 




Chemical Work . . 


1.10-1.35 


.070 


.25 


.60 


.... 


.... 


Chilling I 


Brake Shoes . . . 
roii 


2.00-2.50 
Below 1.00 


.150 


.70 


.70 


2900 


28,000 





198. Wrought iron and steel. Wrought iron is made by burning 
the impurities out of cast iron. In the process the foundry pig iron 
from the blast furnace is first placed in the puddle furnace, where it 
is heated and stirred until the carbon, silicon, and manganese are 
almost entirely burned out. When taken from the furnace, the iron 
is in the form of a pasty ball, which is squeezed until the cinders are 
expelled, after which it is rolled into bars known as muck bars. After 
being reheated it is rolled again, and is then known as merchant lar. 
If a better grade of wrought iron is desired, the merchant bar is 
reheated and rolled again, when it is known as lest iron; if rolled 
again, the quality is still further improved. 

In methods used by the ancients the ore and fuel were placed 
together. This necessitated a pure fuel and did not admit of rapid 
manipulation ; it is still used, however, to obtain wrought iron of a pure 
quality, and in obtaining very fine grades of steel. 



* Proc. Amer. Soc.for Testing Materials, 1904. 



IRON AND STEEL 



287 



Wrought iron is a tough, ductile material showing an elongation of 
from 18 to 30 per cent in 8 in. Its tensile and compressive strength 
at the elastic limit is about 28,000 lb./in. 2 for high-grade wrought 
iron, and about 23,000 lb./in. 2 for common wrought iron. Its max- 
imum tensile strength varies from 44,000 lb./in. 2 to 64,000 lb./in. 2 
The material is much more elastic than cast iron, its modulus of 
elasticity in tension being about 28,000,000 lb./in. 2 , and in shear 
about 10,000,000 lb./in. 2 

Ingot iron. The impurities of wrought iron have almost been elimi- 
nated in a new product known as ingot iron. In the manufacture of 
this material the carbon, manganese, sulphur, and phosphorus are 
nearly all burned out, leaving the product 99.94 per cent pure iron, 
which greatly increases its strength and ductility (see Ey. Age Gazette, 
Vol. 49, p. 574). It does not corrode easily, and has good electrical 
conductivity and low magnetic retentivity. 

199. Manufacture of steel. Tool steel is made by recarbonizing 
wrought iron by heating it in a charcoal fire for several days at a 
temperature of about 3000 F. During this process part of the carbon 
is absorbed by the iron, the product being known as Ulster steel. 
This is then melted and cast into ingots, from which the merchant- 
able bars are rolled or hammered. The two steps in this process are 
usually combined into one. 

Tool steel. Carbon tool steel, such as has been used until within the 
past few years, did not admit of high speeds when cutting. The rub- 
bing of the chip soon dulled the tool, and any considerable increase in 
temperature was sufficient to cause it to lose its hardness. It has been 
unusual for such steel to stand a speed of cutting of 50 ft./min. 

The constituents of carbon steel as previously used for cutting tools 
are indicated by the following table (see Becker, High-Speed Steel) : 



USE 


IRON 


MANGA- 
NESE 


SILICON 


SUL- 
PHUR 


PHOS- 
PHORUS 


CARBON 


Hammers 


99.040 


.21 


.21 


.022 


.020 


.50 to .75 


Knives . 


98 935 


20 


18 


.020 


.015 


.65 to .80 


Drills, reamers . . . 


98.731 


.18 


.21 


.015 


.014 


.85 to 1.30 


Lathe tools .... 


98.520 


.26 


.20 


.010 


.010 


1.00 to 1.30 


Razors 


98.265 


.22 


.20 


.006 


.009 


1.30 to 1.50 


Carving tools . . . 


98.374 


.16 


.14 


.014 


.012 


1.30 to 1.50 



288 



STRENGTH OF MATERIALS 



High-speed steel. Recently it has been found that the addition of 
tungsten and other constituents had the effect of so changing the tool 
steel as to increase its wearing qualities and to make it capable of 
cutting at a much higher speed than formerly. A speed of 500 ft. 
per minute is often obtained with this new steel, although the average 
is considerably less than this. The tool may be heated up to redness 
in cutting without injuring its wearing qualities appreciably. This 
high-speed steel, as. it is called, has made very rapid work possible. 

The chemical analysis of twenty brands of this material is given 
by Becker as follows : 





AVERAGE 


HIGH 


Low 


Carbon 
Tun (r st6ii 


.75 
18 00 


1.28 
25 45 


.32 
14 23 


Molybdenum .... 
Chromium 
Vanadium 


3.50 
4.00 
30 


7.6 

7.2 
32 


0.00 
2.23 
00 


Manganese .... 
Silicon 


.13 
22 


.30 
1 34 


.03 
43 


Phosphorus .... 
Sulphur 


.018 
010 


.029 
016 


.013 

008 











Open-hearth steel is obtained by mixing molten pig iron with scrap iron 
or scrap steel in an open-hearth furnace. The added scrap is low in car- 
bon, and thus lowers the percentage of carbon in the mixture. To offset 
this, the desired amount of carbon is introduced by adding spiegeleisen. 

Bessemer steel is made directly from pig iron in a Bessemer con- 
verter, no additional fuel other than the impurities in the metal 
being used. These impurities are burned out to the desired extent 
by forcing jets of hot air through the liquid metal. Since in this 
method the molten iron is taken directly from the blast furnace, \\ 
considerable saving in the cost of production is effected, by reason of 
which the Bessemer process has revolutionized the steel industry. 

In both the open-hearth and Bessemer processes the liquid steel is 
cast into ingots, which are rolled into the desired shapes. 

200. Composition of steel. The physical properties of steel are 
largely modified by the relative proportions in which the various 
ingredients are present. 



IRON AND STEEL 



289 



Carbon. Increasing the amount of carbon in steel has, in general, the 
effect of increasing its modulus of elasticity and its ultimate strength. 
From a series of tests made on carbon steel, in which the percentage 
of carbon varied from 0.08 to 1.47, Professor Arnold found that the 
elastic limit varied from 27,300 lb./in. 2 to 72,300 lb./in. 2 ; the tensile 
strength, from 47,900 lb./in.' 2 to 124,800 lb./in. 2 ; the elongation, from 
46.6 per cent to 2.80 per cent; and the reduction of area, from 74.8 per- 
cent to 3.30 per cent* The following table gives average values of 
the ultimate strength in both tension and compression for Bessemer 
and open-hearth steel containing different percentages of carbon. 





TENSILE STUESS 


COMPRESSIVE 
STRESS 


SHEARING 

STRESS 


PEK CENT OF 








CARBON 












Elastic Limit 


Maximum 


Elastic Limit 


Maximum 




lb./in. 2 


lb./in. 2 


lb./in.2 


lb./in.2 


0.15 


42,000 


64,000 


40,000 


48,000 


0.20 


46,000 


70,000 


43,000 


54,000 


0.50 


50,000 


78,000 


48,000 


56,000 


0.70 


54,000 


88,000 


54,000 


59,000 


0.80 


58,000 


99,000 


62,000 


68,000 


0.06 


70,000 


115,000 


70,000 


80,000 



Carbon tool steel furnishes material for springs, saws, chisels, files, 
etc. When annealed it is strong in both tension and compression, 
and quite ductile, but when heated to the critical temperature and 
then quenched, it becomes weak, brittle, and hard. 

Silicon in carbon steel and wrought iron generally strengthens the 
material, but decreases its ductility. In amount it is usually less than 
0.6 per cent. 

Manganese increases both the strength and hardness of carbon steel 
and wrought iron, and decreases ductility to some extent. More than 
1.5 per cent makes the steel very brittle. When manganese is present 
in quantities of from 10 to 35 per cent, with a small amount of carbon, 
say 1 per cent, the steel becomes hard and is used for castings and 
forgings. When annealed the castings are both strong and tough 
enough to resist wear. Eolled manganese steel is also produced. 

*Proc. Inst. Civ. Eng., 1895. 



290 



STRENGTH OF MATERIALS 



Manganese steel. This is coming into use for railroad rails on account 
of its resistance to wear. The average values for the strength of this 
material may be given as follows : 





ELASTIC LIMIT 
lb./in. 


TENSILE STRENGTH 
Ib./in.a 


Cast manganese steel 
Rolled manganese steel .... 


45,000 
60,000 


82,000 
135,000 



Sulphur increases the brittleness and hardness of steel and wrought 
iron, and is, in general, a very harmful ingredient. Low percentages 
-of sulphur somewhat increase the tensile strength. 

Phosphorus increases hardness and tensile strength, but decreases 
ductility, making the metal weak under impact and unsuited for any- 
thing but static loads. 

Nickel is added to steel up to about 35 per cent. When the per- 
centage of nickel is low, say about 5 per cent or less, the elastic limit 
and tensile strength are raised without any reduction in the elonga- 
tion or in the contraction of area. Because of this increase in strength 
without loss of ductility, nickel steel is used in the manufacture of 
armor plate, armor-piercing shells, boiler tubes, shafting, etc., where 
a steel is needed which shall combine great strength with toughness. 
The following table shows the relative properties of low carbon steel 
tubes and high nickel steel tubes.* 



PROPERTIES 


Low CARBON STEEL 
TUBES 


HIGH NICKEL STEEL 
TUBES 


Tensile strength, lb./in. 2 
Elastic limit, lb./in. 2 . . . 
Per cent of elongation in 8 in. 


60,000-55,000 
30,000-35,000 
20-30 


85,000-95,000 
40,000-45,000 
20-30 



The same authority also gives the average tensile strength of six- 
teen steel tubes, composed of 25 per cent nickel, as 108,913 lb./in. 2 
for unannealed specimens, and 97,300 lb./in. 2 for annealed specimens. 
The elongation in the former case was 28 per cent in 7.87 in., and in 
the latter case 38 per cent in the same length. 

* Proc. Soc. Naval Architects and Marine Engineers, November, 1903. 



IRON AND STEEL 291 

Tests made by the Watertown Arsenal on a 3.37 per cent nickel 
steel gave an average elastic limit of 56,700 lb./in. 2 and a tensile 
strength of 90,300 lb./in. 2 * 

Vanadium steel. Vanadium when added to steel in small quantities 
acts as a dynamic intensifier, that is to say, it greatly increases resist- 
ance to fatigue under alternating or repeated stresses. Vanadium 
also increases the static properties of steel, increasing its strength 
and toughness and resistance to wear or abrasion. Tensile tests 
made by the writer on one grade gave the following results: Elastic 
limit, 78,000 lb./in. 2 ; yield point, 91,900 lb./in. 2 ; tensile strength, 
116,100 lb./in. 2 ; modulus of elasticity, 30,900,000 lb./in. 2 ; elonga- 
tion in 8 in., 20 per cent; reduction of area at fracture, 57 per cent. 
This material showed the following chemical analysis : Vanadium, 
.30; carbon, .25; manganese, .15; chrome, .42; phosphorus, .009; 
sulphur, .024; silicon, .10. 

Recent tensile tests of large I-bars of vanadium steel having a 
cross section of 14 in. x 2 in. gave an average elastic limit of 70,000 
lb./in. 2 , an average yield point of 81,200 lb./in. 2 , and an average 
maximum strength of 96,800 lb./in. 2 (see Eng. Record, July 30, 
1910). The following chemical analysis is given for this steel: car- 
bon, .25 ; vanadium, .17 ; nickel, 1.45 ; chrome, 1.20 ; manganese, .32 ; 
silicon, .12; phosphorus, .02; sulphur, .035. In engineering steels 
the maximum amount required seldom exceeds 0.2 per cent. Its 
judicious use makes it possible to fulfill varied requirements, whether 
chiefly static, chiefly dynamic, or divided between the two. 

201. Steel castings are made both by the Bessemer and open- 
hearth processes. In the Bessemer process the iron is first reduced 
to wrought iron, and then spiegeleisen, or ferromanganese, added to 
furnish the necessary carbon. Aluminum may be added to pre- 
vent blowholes. The metal is cast in the same way as in making 
other castings. 

On page 292 is given a report of a series of tests made at the 
Watertown Arsenal on castings for gun carriages.! The elastic limit 
varied from 47,000 lb./in. 2 to 21,500 lb./in. 2 , and the tensile strength 
from 81,000 lb./in. 2 to 43,000 lb./in. 2 Good average values might be 
given as 30,000 lb./in. 2 at the elastic limit and 66,000 lb./in. 2 at the 

* Watertown Arsenal Report, 1899. t Watertown Arsenal Report, 1903. 



292 



STRENGTH OF MATERIALS 



maximum. At the elastic limit the compressive strength was about 
the same as the tensile strength. The American Society for Testing 



TEST OF STEEL CASTINGS 



ELASTIC 
IJMIT 

lb./in. 


TENSILE 
STRENGTH 

Ib./in.* 


ELONGA- 
TION IN 

2 IN. 

per cent 


CONTRAC- 
TION OF 
AREA 

per cent 


APPEARANCE OF FRACTURE 


43,500 


81,500 


24.5 


49.1 


Fine silky 


43,500 


78,600 


28.5 


49.1 


" 


44,000 


80,500 


26.5 


46.2 


" 


47,000 


73,000 


28.5 


59.8 


" 


44,500 


73,500 


28.5 


57.2 


" 


42,000 


78,550 


26.5 


51.9 


< 


46,000 


74,800 


32.0 


59.8 


< 


46,500 


73,600 


28.5 


57.2 


<( 


43,750 


75,200 


30.0 


57.2 


t< 


42,500 


75,100 


28.5 


57.2 


" 


34,000 


67,000 


32.0 


51.9 


Silky 


42,000 


67,100 


32.0 


51.9 


" 


38,500 


67,700 


27.5 


40.2 


" 


26,000 


43,000 


8.5 


20.5 


Dull silky ; granular spots ; blowhole 


25,980 


60,620 


29.0 


42 2 


Dull silky, 80 per cent ; granular, 20 per cent 


24,960 


60,370 


23.5 


29.4 


Granular, silvery luster, 90 per cent ; dull silky, 










10 per cent 


30,060 


60,110 


29.0 


48.1 


Dull silky 


32,500 


68,000 


19.5 


40.3 


Granular, 50 per cent ; dull silky, 50 per cent 


31,500 


66,750 


29.5 


40.3 


Dull silky 


29,000 


66,500 


24.0 


40.3 


ii 


24,450 


60,620 


27.0 


36.0 


Dull silky, 40 per cent ; granular, CO per cent 


27,000 


60,750 


6.5 


16.9 


Dull silky ; blowhole 


26,000 


66,250 


12.5 


23.9 


Granular, silvery luster, 85 per cent ; dull silky, 










15 per cent 


22,500 


58,500 


31.0 


40.3 


Granular, silvery luster, 60 per cent ; dull silky, 










40 per cent 


26,500 


66,750 


20.0 


43.3 


Dull silky 


38,000 


66,250 


22.5 


37.1 


Dull silky, GO per cent ; granular, 40 per cent 


26,000 


69,000 


9.0 


27.4 


Dull silky ; granular spots 


24,450 


59,600 


25.5 


32.8 


Dull silky, 80 per cent ; granular, 20 per cent 


27,500 


60,500 


27.0 


37.1 


Dull silky ; trace of granulation 


25,500 


63,750 


28.5 


37.1 


Dull silky ; granular spots 


31,000 


66,750 


16.5 


16.9 


Granular, silvery luster, 80 per cent ; dull silky, 










20 per cent 


37,500 


68,000 


14.5 


16.9 


Granular, silvery luster, 85 per cent ; dull silky, 










15 per cent 


26,000 


59,000 


21.5 


37.1 


Dull silky ; granular spots 


27,500 


60,500 


21.5 


40.3 


Dull silky, 90 per cent ; granular, 10 per cent 


24,960 


60,880 


26.5 


36.0 


Dull silky 


26,490 


64,700 


17.0 


22.5 


Granular, silvery luster 


25,470 


63,930 


15.5 


19.0 


" 


29,040 


65,210 


29.0 


48.1 


Dull silky 


26,500 


64,500 


30.0 


46.2 


<( 


27,500 


67,750 


15.0 


16.9 


Dull silky, 50 per cent ; granular, 50 per cent 


27,000 


64,750 


16.0 


16.9 


Dull silky, 20 per cent ; granular, 80 per cent 



IRON AND STEEL 



293 



Materials has recommended the following values for the strength of 
steel castings (allowable variation 5000 pounds). TENSILF STRENGTH 

lb./in.2 

Soft castings 60,000 

Medium castings 70,000 

Hard castings 80,000 

In the cold bending test the material must be bent about a diameter 
of 1 iii. through 120 for the soft, and 90 for the medium, without 
showing cracks or signs of failure.* 

The Ordnance Department of the United States Army in the gen- 
eral specifications for 1903 gives the following requirements for steel 
castings and forgings. 



METAL 


ELASTIC 
LIMIT 

Ib./in.* 


TENSILE 
STRENGTH 

Ib./in.z 


ELONGATION 
AFTER RUP- 
TURE 

per cent 


CONTRAC- 
TION OF 
AREA 

per cent 


Cast steel, No. 1 ... 


( 25,000 
\ 28,000 


60,000 
65,000 


18.0 
16.0 


27.0 
24.0 


Cast steel, No. 2 ... 


35,000 


75,000 


15.0 


20.0 


Cast steel, No. 3 ... 


45,000 


85,000 


12.0 


18.0 


Forged steel, No. 1 ... 


27,000 


60,000 


28.0 


40.0 


Forged steel, No. 2 . . . 


35,000 


75,000 


20.0 


30.0 


Forged steel, No. 3 ... 


42,000 


90,000 


16.0 


24.0 



202. Modulus of elasticity of steel and wrought iron. The mod- 
ulus of elasticity of steel and wrought iron is about the same in 
tension as in compression. For steel, 30,000,000 lb./in. 2 is usually 
taken as a good average value for tension and compression, and 
about two fifths of this amount, or from 10,000,000 to 12,000,000 
lb./in. 2 , for shear; for loads below the elastic limit it is always the 
ratio of stress to deformation. 

From a series of tests reported in the Trans. Amer. Soc. Civ. Eng., 
Vol. XVII, pp. 62-63, the following average values are found. 



MODULUS OF ELASTICITY 



MATERIAL 





Tension 


Compression 


Ordinary steel 


30 000 000 


29,000,000 


Spring steel . . . . 


29 500 000 


29,300,000 


Wrought iron 


28 200 000 


27 600,000 









* Proc. Amer. Soc. for Testing Materials, 1903. 



294 



STRENGTH OF MATERIALS 



203. Standard form of test specimens. It was pointed out in 
Article 20 that the form of the test specimen had considerable effect 
upon the results obtained from tests. To eliminate this factor, standard 



*/ 

About -3 A N/.g 


Parallel section not less than 9" 


[-About 3^-* 






*t 




Csj 




r t t *T* 




T 1 




$r-&r-&Elc. 








FIG. 177 

dimensions for both cylindrical and rectangular test specimens have 
been adopted. These are shown in Fig. 177. 

204. Specifications for wrought iron and steel. In order that the 
student may form some idea of the strength required by manufac- 
turers for different grades of wrought iron and steel, quotations are 
given below from the specifications of the American Society for 
Testing Materials. 

WROUGHT IRON 





STAY-BOLT 


MERCHANT 


MERCHANT 


MERCHANT 




IRON 


GRADE A 


GRADE B 


GRADE C 


Tensile strength, lb./in. 2 . . 


46,000 


50,000 


48,000 


48,000 


Yield point, lb./in. 2 .... 


25,000 


25,000 


25,000 


25,000 


Per cent of elongation in 8 in. 


28 


25 


20 


20 



STEEL 





RIVET STEEL 


SOFT STEEL 


MEDIUM STEEL 


Tensile strength, lb./in. 2 . . 


50,000-60,000 


62,000-62,000 


60,000-70,000 


Yield point, lb./in. 2 . . . . 


30,000 


32,000 


35,000 


Elongation in per cent for 8 in. 








shall not be less than . . 


26 


25 


22 



IKON AND STEEL 



295 



The above grades of steel, known as structural steel for bridges and ships, 
must conform to certain bending tests. For this purpose the test specimens 
shall be 1| in. wide, if possible, and for all material f in. or less in thickness the 
test specimen shall be of the same thickness as that of the finished material 
from which it is cut ; but for material more than in. thick the bending test 
specimen may be \ in. thick. Rivet rounds shall be tested full size as rolled. 

Rivet steel shall bend cold 180 flat on itself without fracture on the out- 
side of the bent portion. 

Soft steel shall bend cold 180 flat on itself without fracture on the outside 
of the bent portion. 

Medium steel shall bend cold 180 around a diameter equal to the thick- 
ness of the specimen tested, without fracture on the outside of the bent 
portion. 

STEEL AXLES 

Steel for axles shall be made by the open-hearth process and shall be 
divided into the following classes : (a) car, engine-truck, and tender-truck 
axles ; and (J) driving axles. For (a) no tensile tests shall be required, but 
for driving axles the following physical properties shall be required. 





CARBON STEEL 


NICKEL STEEL 


Tensile strength Ib./in 2 


80 000 


80 000 


Yield point, Ib./in. 2 


40 000 


50 000 


Contraction of area in per cent .... 




45 


Per cent of elongation in 2 in 


20 


25 



The same specifications require that one axle taken from each 
melt shall be tested by the drop test, as follows. 



DIAMETER OF AXLE AT 
CENTER 
in. 


NUMBER OF BLOWS 


HEIGHT OF DROP 
ft. 


DEFLECTION 
in. 


*J 


5 


24 


8 i 


4 f 


5 


26 


81 


4jV 


5 


28 ' 


81 


4f 


5 


31 


8 


4| 


5 


34 


8 


5| 


5 


43 


7 


5! 


7 


43 


5* 



To be accepted, the axle must stand the blow without rupture and without 
exceeding, as the result of the first blow, the deflection stated. 



296 STRENGTH OF MATERIALS 

DESCRIPTION OF THE DROP TEST 

The points of support on which the axle rests during tests shall be 3 ft. 
apart from center to center; the hammer must weigh 1640 Ib. ; the anvil, 
which is supported on springs, must weigh 17,500 Ib. ; it must be free to 
move in a vertical direction ; the springs upon which it rests must be twelve 
in number, of the kind specified ; and the radius of supports and of the strik- 
ing face on the hammer in the direction of the axis of the axle must be 5 in. 

The deflections are measured by placing a straightedge along the 
axle, properly held at the supports, and measuring the distance from 
this straightedge to the axle both before and after the blow. The 
difference between the two measurements gives the deflection. 



CHAPTER XIII 

LIME, CEMENT, AND CONCRETE 

205. Quicklime. If calcium carbonate (ordinary limestone) is 
heated to about 800 F., carbon dioxide is driven off, leaving an 
oxide of calcium, which is known as quicklime. This has a great 
affinity for water and slacks upon exposure to moisture. Slacked 
lime when dry falls into a fine powder. 

Lime mortar is formed by mixing slacked lime with a large propor- 
tion of sand. Upon exposure to the air this mortar becomes hard by 
reason of the lime combining with carbon dioxide and forming again 
calcium carbonate, the product being a sandy limestone. Lime mortar 
is used in laying brick walls and in structures where the mortar will 
not be exposed to water, since it will not set, i.e. combine with carbon 
dioxide, under water. 

206. Cement. When limestone contains a considerable amount of 
clay, the lime produced is called hydraulic lime, for the reason that 
mortar made by using it will harden under water. If the limestone 
contains about 30 per cent of clay and is heated to 1000 F., the 
carbon dioxide is driven off, and the resulting product, when finely 
ground, is called natural cement. When about 25 per cent of water is 
added, this cement hardens, because of the formation of crystals of 
calcium and aluminum compounds. 

If Limestone and clay are mixed in the proper proportions, usually 
about three parts of lime carbonate to one of clay, and the mixture 
roasted to a clinker by raising it to a temperature approaching 3000 F., 
the product, when ground to a fine powder, is known as Portland cement 
The proper proportion of limestone and clay is determined by find- 
ing the proportions of the particular clay and stone that will make 
perfect crystallization possible. In the case of natural cement the 
lime and clay are not present in such proportions as to form perfect 
crystals, and consequently it i^ not as strong as Portland cement. 

297 



298 STRENGTH OF MATERIALS 

The artificial mixing of the limestone and clay in the manufacture 
of Portland cement is accomplished in different ways. Throughout 
the north central portion of the United States large beds of marl are 
found, and also in the same localities beds of suitable clay. This marl 
is nearly pure limestone, and is mixed with the clay wet. (These 
materials are also mixed dry.) Both the marl and clay are pumped to 
the mixer, where they are mixed in the proper proportions. The prod- 
uct is then dried, roasted, and ground. Most American Portland ce- 
ments, however, are made by grinding a clay-bearing limestone with 
sufficient pure limestone to give the proper proportions. After being 
thoroughly mixed the product is roasted and ground to a powder. 

Slag cement (Puzzolan) is made by thoroughly mixing the granulated 
slag from an iron blast furnace with slacked lime, and then grinding 
the mixture to a fine powder. Slag cements are usually lighter in 
color than the Portland cements, and have a lower specific gravity, the 
latter ranging from 2.7 to 2.8. They are also somewhat slower in 
setting than the Portland cements, and have a slightly lower tensile 
strength. They are not adapted to resist mechanical wear, such as 
would be necessary in pavements and floors, but are suitable for 
foundations or any work not exposed to dry air or great strain. 

True Portland cement may be made from a mixture of blast-furnace 
slag and finely powdered limestone, the mixture being burned in a 
kiln and the resultant clinker ground to powder. Both the Portland 
and the Puzzolan cements will set under water, i.e. they are hydraulic. 

207. Cement tests. The many different processes of mixing, roast- 
ing, grinding, and setting through which a cement must pass, require 
that a number of tests be made to determine whether or not these 
have been well done. If the grinding has been improperly done, or 
if any of the other operations of manufacture have been neglected, 
the product may be very weak, or even worthless. To make sure that 
all the steps in the manufacture of the cement have been properly 
carried out, engineers make use of the following tests : (a) test of 
soundness ; (6) test of fineness ; (c) test of time of setting ; (d) test 
of tensile strength. 

208. Test of soundness. One test for soundness consists in boiling 
a small ball of neat cement in water for three hours, and noting 
whether or not checks or cracks occur. If the cement contains too 



LIME, CEMENT, AND CONCRETE 299 

much free lime, the ball will disintegrate and show signs of crumbling. 
The ball of cement is kept under a damp cloth for twenty-four 
hours before boiling. This test is not regarded with favor by many 
engineers (see steam pat test, specifications, p. 305). 

209. Test of fineness. If the grinding has not been properly done, 
large particles of clinker remain, which act as a sand or other foreign 
substance and thus weaken the cement. The test for fineness is made 
by sifting the cement through different sieves ; usually all of it is 
required to pass through a sieve of 50 meshes to the inch, and a 
smaller amount through sieves of 80 and 100 meshes. About 75 per 
cent should pass through a 200-mesh sieve (see Article 214). 

210. Test of time of setting. It is important that a cement should 
not set too quickly or too slowly. A test for time of setting, known 
as Grillmore's test, has been standardized in the United States, and 
consists in applying to a small cement pat given weights supported 
by points of specified area (Fig. 178). The cement pat is made by 
mixing a portion of neat cement with the proper amount of water, 
mounting this on a piece of glass, and smoothing it until the middle 
is half an inch thick and the edges are smooth and tapering. The 
pat is then kept under a damp cloth to prevent injury by sudden 
changes in temperature, or too high temperature, of the surrounding 
air. When this pat will hold without appreciable indentation a 
quarter-pound weight supported by a wire ^ in. in diameter, it is 
said to have acquired its initial set. It is said to have acquired its 
final set when a one-pound weight supported by a wire ^ in. in 
diameter will not appreciably indent the surface. 

When a pat prepared as indicated above checks or warps, it 
indicates that the cement in setting changes volume too rapidly. 
For many pieces of work a slow-setting cement cannot be used ; but 
a cement which sets too quickly is likely to contain too much free 
lime, and should be very carefully tested before being used. In 
general, the time of final setting for natural cement should not be 
less than thirty minutes nor more than three hours. 

The table given on page 300 shows the time of setting of different 
brands of cement.* The student is also referred to the standard speci- 
fications for cement given in Article 214. 

*Watertown Arsenal Report, 1901. 



300 



STRENGTH OF MATERIALS 



TIME OF SETTING OF CEMENTS 







TIME OF SETTING 


BRAND OF 


W 
H 


Gillmore's Method 


German Method 


CEMENT 


[ 










Initial 


Final 


Interval 


Initial 


Final 


Interval 




Percent 


Hr. Min. 


Hr. Min. 


Hr. Min. 


Hr. Min. 


Hr. Min. 


Hr. Min. 


Alpha .... 


20 


2 20 


5 00 


2 40 


35 


4 25 


3 50 




- 25 


3 20 


7 30 


4 10 


2 50 


6 35 


3 45 




30 


5 40 






4 40 


8 40 


4 00 


Atlas 


20 


4 05 


7 10 


3 05 


2 45. 


6 10 


3 25 




25 


5 10 


8 05 


2 55 


3 35 


7 05 


3 30 




30 


7 00 






5 30 






Star, with plaster 


20 


2 10 


4 25 


2 15 


50 


3 00 


2 10 




25 


4 35 


6 00 


1 25 


3 00 


5 30 


2 30 




30 


5 45 






5 10 


7 15 


2 05 


S t a r, w i t h o u t 
















plaster . . . 


20 


05 


15 


10 


05 


10 


05 




25 


35 


4 55 


4 20 


10 


3 30 


3 20 




30 


5 10 


8 35 


3 25 


3 15 


6 50 


3 35 


Whitehall . . . 


20 


1 49 


5 19 


3 30 


1 28 


4 44 


3 16 




25 


4 15 


6 05 


1 50 


3 25 


5 40 


2 15 




30 


4 59 


7 19 


2 20 


4 33 


6 53 


2 20 


Josson .... 


20 


30 


4 35 


4 05 


05 


3 40 


3 35 




25 


4 10 


6 40 


2 30 


3 10 


6 10 


3 00 




30 


5 35 


8 05 


2 30 


4 50 


7 20 


2 30 


Storm King . . 


25 


4 02 


6 57 


2 55 


1 42 


5 37 


3 55 




28 


5 30 






4 20 


7 05 


2 45 




30 


5 47 






4 27 






Alsen 


25 


25 


'l 'is' 


'o 'so' 


10 


'o '35' 


'o '25' 




30 


30 


1 15 


45 


20 


45 


25 




35 


2 30 


4 00 


1 30 


30 


1 50 


1 20 


Silica . . 


20 


20 


2 52 


2 32 


13 


37 


24 




25 


29 


4 59 


4 30 


22 


1 49 


1 27 


Cathedral . . . 


22 


4 52 


6 17 


1 25 


1 12 


5 32 


4 20 




25 


4 45 


6 55 


2 10 


2 40 


6 10 


3 30 




28 


5 05 


7 29 


2 24 


3 33 


6 45 


3 12 


Akron Star . . 


30 


2 25 


6 30 


4 05' 


45 


4 55 


4 10 




35 


4 05 


7 10 


3 05 


2 45 


6 35 


3 50 




40 


6 55 






6 15 






Austin . . . 


30 


47 


2 51 


2 04 


16 


2 08 


1 52 




35 


1 03 


3 18 


2 15 


43 


2 28 


1 45 




40 


1 23 


4 48 


3 25 


1 08 


3 58 


2 50 


Hoffman . . . 


30 


2 15 


3 25 


1 10 


1 25 


2 55 


1 30 




35 


2 55 


5 40 


2 45 


2 20 


4 10 


1 50 




40 


3 43 






2 48 






Norton .... 


30 


37 


2 12 


1 35 


25 


1 00 


35 




35 


49 


3 14 


2 25 


34 


1 54 


1 20 




40 


1 02 


5 17 


4 15 


40 


3 37 


2 57 


Obelisk .... 


30 


1 40 


4 05 


2 25 


25 


3 20 


2 55 




35 


2 47 


5 02 


2 15 


1 49 


4 12 


2 23 




40 


3 15 


5 20 


2 05 


2 50 


4 15 


1 25 


Potomac . . . 


30 


45 


3 40 


2 55 


25 


1 55 


1 30 




35 


1 05 


4 43 


3 38 


43 


2 58 


2 15 


Newark and Ros- 


40 


1 15 


5 30 


4 15 


1 07 


4 25 


3 18 


endale . . . 


35 


37 


1 17 


40 


32 


1 07 


35 




40 


47 


3 44 


2 57 


40 


2 19 


1 39 




45 


1 08 


4 18 


3 10 


48 


3 33 


2 45 


Mankato . . . 


40 


2 40 


5 15 


2 35 


1 20 


3 55 


2 35 




45 


2 59 






2 29 


4 19 


1 50 




50 


3 24 






2 50 


5 09 


2 19 







211. Test of tensile strength. The tensile strength of a cement is 
made by testing briquettes of neat cement or cement mortar in 
tension. The briquettes are made in standard molds (Fig. 180), 




FIG. 178. Weights for Testing Briquettes 




EIG. 170. Cement Testing Machine 



LIME, CEMENT, AND. CONCRETE 



301 




FIG. 180 



which provide for a cross section of one square inch at the middle, 
with thicker ends for insertion in the jaws of the testing machine. 
This test requires considerable expertness to get satisfactory results, 
for the proper mix- 
ing and tamping into 
the molds can only 
be satisfactorily done 
by one of consider- 
able experience. After molding, the briquettes are kept under a damp 
cloth for about twenty-four hours and then under water until tested. 

Many machines are now made for testing the tensile strength of 
cement, most of them being light enough to be portable. A new 
automatic machine, manufactured by the Olsen Testing Machine 
Company of Philadelphia, is shown in Fig. 179. The machine is 
operated by first placing the briquette in position and balancing the 
beam at the top. The load is then applied by allowing the shot to run 
from the pan on the right end of the beam. The spring balance gives 
the exact weight of the shot and, consequently, the tensile stress on the 
briquette at any time during the test. After the briquette is broken 
the tensile strength in pounds per square inch is recorded on the dial. 

212. Speed of application of load. It has been found that the 
rapidity with which the load is applied has considerable effect upon 
the results obtained in making tension tests of cement. The follow- 
ing table clearly shows this effect.* 



EFFECT OF SPEED OF APPLICATION OF LOAD ON TENSILE 
STRENGTH OF CEMENT 





SPEED 






SPEED 




NUMBER 


IN POUNDS- 
SECONDS 


AVERAGE 


NUMBER 


IN POUNDS- 
SECONDS 


AVERAGE 


OF 

BRIQUETTES 




RESULTS 


BRIQUETTES 




RESULTS 












Ib. 


sec. 






Ib. 


sec. 




129 


100 


1 


5.60.75 


90 


100 


30 


417.27 


129 


100 


15 


606.43 


90 


100 


00 


403.00 


145 


100 


15 


452.2 


40 


100 


60 


416.75 


145 


100 


30 


430.96 


40 


100 


120 


400.00 



* Proc. Inst. Civ. Eng., 1883. 



302 



STRENGTH OF MATERIALS 



TENSILE AND COMPRESSIVE TESTS OF CEMENT 



BRAND OF 

CEMENT 


TENSILE TEST 


COMPRESSION TEST 


Age 


Ib./in.z 


Sectional 
Area 
in. 2 


Age 


Total 
Ib. 


lb./in.2 


Air 
(days) 


Water 
(days) 


Months 


Days 


Atlas .... 




6 


1066 


2.80 


1 


20 


22,050 


7,875 






G 


1012 


2.83 


1 


20 


33,700 


11,908 






6 


957 


2.89 


1 


20 


33,600 


11,626 






6 


775 


2.78 


1 


20 


31,020 


11,158 






6 


759 


2.35 


1 


19 


28,100 


11,957 




1 


6 


738 


2.90 


1 


8 


25,800 


8,896 




1 


6 


698 


2.82 


1 


8 


25,900 


9,184 




1 


6 


654 


2.46 


1 


8 


19,500 


7,927 




1 


6 


615 


2.92 


1 


8 


26,100 


8,938 


Storm King . 


1 


6 


174 


2.27 


1 


15 


8,300 


3,656 




1 


6 


331 


2.56 


1 


15 


10,620 


4,148 




1 


6 


354 


2.37 


1 


15 


8,100 


3,418 




1 


6 


189 


2.24 


1 


14 


8,640 


3,857 




7 




372 


2.53 


1 


14 


7,560 


2,988 




7 




443 


2.16 


1 


14 


7,020 


3,250 




7 


. . . 


543 


2.26 


1 


14 


7,540 


3,336 


Alsen .... 


1 


6 


714 


2.26 


. 


24 


12,500 


5,531 




1 


6 


670 


2.36 


. 


24 


13,300 


5,635 




1 


6 


755 


2.91 


. 


23 


16,500 


5,670 




7 




446 


2.53 




23 


13,270 


5,245 




7 




391 


2.58 




21 


11,300 


4,380 


Dyckerhoff . . 


7 




471 


2.54 


1 


8 


13,900 


5,472 




7 




515 


2.65 


1 


8 


16,300 


6,151 




7 




226 


2.73 


1 


7 


14,400 


5,275 




7 




279 


2.56 


1 


7 


14,300 


5,586 


Steel .... 


1 


6 


298 


2.22 


. . . 


13 


3,840 


1,730 




1 


6 


332 


2.67 


. . . 


13 


5,300 


1,985 




7 




99 


2.37 


. 


13 


4,720 


1,991 




7 




149 


2.48 


. 


13 


4,520 


1,822 




7 


. . . 


139 


2.20 




13 


4,100 


1,864 


Bonneville . . 


1 


6 


196 


2.39 




15 


4,100 


1,715 


Improved . . 


1 


6 


158 


2.10 




15 


3,180 


1,514 




1 


6 


42 


2.22 




15 


3,120 


1,405 




1 


6 


81 


2.22 




15 


3,020 


1,360 




7 




201 


2.14 




14 


2,710 


1,266 




7 




73 


2.24 




14 


2,710 


1,209 




7 




196 


2.11 




14 


2,400 


1,137 




7 




99 


2.03 




14 


2,520 


1,241 


Hoffman . . . 


1 


6 


156 


2.48 


. 


21 


5,060 


2,040 




1 


6 


133 


2.41 


. 


21 


4,920 


2,041 




7 


. . . 


60 


1.94 




20 


3,030 


1,562 




7 


. 


53 


2.22 




20 


3,890 


1,752 




7 


. 


243 


2.56 




20 


4,660 


1,820 




7 




225 


2.35 




20 


4,550 


1,936 




7 




275 


2.28 




20 


4,420 


1,939 



LIME, CEMENT, AND CONCRETE 303 

213. Compression tests. Compression tests of cement are made 
in Europe, but not generally by engineers in the United States, as 
the tensile test is thought quite as valuable as the compression test 
in giving results indicative of the strength of the cement. Compres- 
sion tests are made upon the ends of the specimen broken in tension, 
or upon specially prepared cement cubes. The use of the broken 
ends of the briquette insures the same material for the compression 
test as was used in the tension test. The table on page 302 gives the 
compressive strength of several brands of cement.* The tests were 
made by compressing halves of briquettes broken in tension., and 
both the tensile and compressive strengths are given. 

214. Standard specifications for cement. The following is a copy 
of the standard specifications for cement adopted by the American 
Society for Testing Materials. 

NATURAL CEMENT 

This term shall be applied to the finely pulverized product resulting from 
the calcination of an argillaceous limestone at a temperature only sufficient 
to drive off the carbonic acid gas. 

Fineness. It shall leave by weight a residue of not more than 10 per cent 
on the No. 100 sieve, and not more than 30 per cent on the No. 200 sieve. 

Time of setting. It shall develop initial set in not less than ten minutes, 
and hard set in not less than thirty minutes nor more than three hours. 

Tensile strength. The minimum requirements for tensile strength forjbri- 
quettes 1 in. square in cross section shall be as follows, and shall show no 
retrogression in strength within the periods specified. 



Neat Cement 

AGE STRENGTH 

24 hours in moist air 75 Ib. 

7 days (1 day in moist air, 6 days in water) 150 " 

28 days (1 27 " ) 250 



One Part Cement, Three Parts Standard Sand 

7 days (1 day in moist air, 6 days in water) ..... 50 Ib. 
28 days (1 27 ) 125 

* Watertown Arsenal Report, 1901. 



304 STRENGTH OF MATERIALS 

Constancy of volume. Pats of neat cement about 3 in. in diameter, | in. 
thick at the center, tapering to a thin edge, shall be kept in moist air for a 
period of twenty-four hours. 

(a) A pat is then kept in air at normal temperature. 

(ft) Another is kept in water maintained as near 70 F. as practicable. 

These pats are observed at intervals for at least twenty-eight days, and, to 
satisfactorily pass the tests, should remain firm and hard and show no signs 
of distortion, checking, cracking, or disintegrating. 



PORTLAND CEMENT . 

This term is applied to the finely pulverized product resulting from the 
calcination to incipient fusion of an intimate mixture of properly proportioned 
argillaceous and calcareous materials, and to which no addition greater than 
3 per cent has been made subsequent to calcination. 

Specific gravity. The specific gravity of the cement, thoroughly dried at 
100 C., shall be not less than 3.10. 

Fineness. It shall leave by weight a residue of not more than 8 per cent on 
the No. 100 sieve, and not more than 25 per cent on the No. 200 sieve. 

Time of setting. It shall develop initial set in not less than thirty minutes, 
and hard set in not less than one hour nor more than ten hours. 

Tensile strength. The minimum requirements for tensile strength for bri- 
quettes 1 in. square in section shall be as follows, and shall show no retro- 
gression in strength within the periods specified. 

Neat Cement 
AGE STRENGTH 

* 24 hours in moist air 175 Ih. 

7 days (1 day in moist air, 6 days in water) 500 " 

28 days (1 27 " ) 600 " 

One Part Cement, Three Parts Standard Sand 

7 days (1 day in moist air, 6 days in water) 200 Ib. 

28 days (1 " " 27 " " ) 275 

Constancy of volume. Pats of neat cement about 3 in. in diameter, ^ in. 
thick at the center, and tapering to a thin edge, shall be kept in moist air 
for a period of twenty-four hours. 

(a) A pat is then kept in air at normal temperature and observed at intervals 
for at least twenty-eight days. 

(5) Another pat is kept in water maintained as near 70 F. as practicable, 
and observed at intervals for at least twenty-eight days. 



LIME, CEMENT, AND CONCRETE 305 

(c) A third pat is exposed in any convenient way in an atmosphere of 
steam, above boiling water, in a loosely closed vessel for five hours. 

These pats, to satisfactorily pass the requirements, shall remain firm and 
hard and show no signs of distortion, checking, cracking, or disintegrating. 

Sulphuric acid and magnesia. The cement shall not contain more than 1.75 
per cent of anhydrous sulphuric acid (SO 3 ), nor more than 4 per cent of 
magnesia (MgO). 

215. Concrete. When cement mortar is mixed with certain per- 
centages of broken stone, gravel, or cinders, the mixture is called 
concrete. The amount and kind of stone or other material to be used 
depends upon the use to be made of the finished product. Concrete 
is rapidly coming into favor as a building material, and is replacing 
brick and stone in many classes of structures. If properly made it is 
a much better building material than either of the latter, and has an 
additional advantage in the fact that it can be handled by unskilled 
labor and may be readily molded into any desired form. In view of 
these facts, a study of its properties is of the greatest importance. 

216. Mixing of concrete. In making concrete, the sand and cement 
are first thoroughly mixed and gauged with the right amount of water. 
The stone, having previously been moistened, is then added, and the 
whole is thoroughly mixed until each piece of stone is coated with the 
cement mortar. These two operations are often combined into one. 
The amount of water to be used in making the mortar depends upon 
the character of the concrete desired. A medium concrete may be 
obtained by adding enough water so that moisture comes to the 
surface when the mortar is struck with a shovel. 

After mixing, the concrete is tamped, or rammed, into position. 
This tamping should be thoroughly done, since in no other way can 
as dense a mixture be obtained. It is desirable that all the voids 
(spaces between the broken stone) should be filled as compactly as 
possible with mortar. 

217. Tests of concrete. Concrete is usually tested in compression, 
and for this purpose 6-inch cubes * are made, composed of cement, 
sand, and broken stone in the proportions of 1 : 2 : 4 or 1 : 3 : 6. In 
some cases the proportion to be used in the particular work con- 
cerned is also used in making the test cubes. These cubes are made 
in molds and allowed to set in air, or part of the time in air and the 

* Cylinders or larger cubes are also sometimes used. 



306 STRENGTH OF MATERIALS 

remainder in water, until tested. The kind of cement as well as its 
physical properties must be known ; also the kind of sand and stone 
and the degree of fineness of each. 

When ready for testing, the concrete cubes are placed in the testing 
machine, bedded with plaster of Paris or thick paper, and tested in 
compression. The load at first crack and the maximum load are noted. 

The table on the opposite page is a report of a series of tests made 
at the "Watertown Arsenal on Akron Star cement concrete in com- 
pression.* It will be noticed that the ultimate strength varied from 
600 lb./in. 2 to 2700 lb./in. 2 

The table on page 308 is taken from the same volume as the 
preceding, and summarizes the results of tests on concrete made 
from different kinds of cement. Various kinds of broken stone were 
used, including broken brick, and the ultimate strength ranged from 
600 lb./iu. 2 to 3800 lb./in. 2 In making comparisons from the table 
as to strength several things must be noted, namely, the kind and 
strength of the cement, the proportions and character of the sand 
and gravel, the treatment after making, and the age when tested ; in 
other words, a complete history of the materials and their treatment 
should be known. In the following table the cubes tested were set 
in air, in a dry, cool place. 

The location and character of the structure will often determine 
the kind of materials to be used in making the concrete. Thus, on 
account of convenience, pebbles are sometimes used with the sand 
in which they are found. This reduces the cost of the concrete, but 
usually impairs its strength, as the proportions of sand and stone as 
they occur in nature are not likely to be such as to be suitable for 
concrete. Theoretically, to get the best results the proportions should 
be such that the cement fills the spaces between the grains of sand, 
and the mortar fills the spaces between the pieces of stone. 

In any particular case the cost of material, strength of the concrete, 
and service required of the structure must determine what propor- 
tions shall be used. 

Problem 300. A concrete cube 12 in. high when tested in compression sustained 
a load of 324,000 Ib. at first crack, and 445,200 Ib. at failure. Find the intensity 
of the compressive stress in lb./in. 2 at first crack and at failure. 

* Watertown Arsenal Report, 1901. 



LIME, CEMENT, AND CONCRETE 



307 



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308 



STRENGTH OF MATERIALS 






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LIME, CEMENT, AND CONCRETE 



309 



218. Modulus of elasticity of concrete. Concrete is so imperfectly 
elastic that the modulus of elasticity varies with the stress. It also 
changes with the age of the material and with the change in propor- 
tions of cement, sand, and stone. 

The variation in the modulus of elasticity with the stress makes 
it difficult to make theoretical computations in which the modulus 
of elasticity is involved, as, for instance, in such problems as arise 
in connection with reenforced concrete beams, etc.* 

MODULUS OF ELASTICITY OF CONCRETE IN COMPRESSION 



COMPOSITION 


AGE 


MODULUS OF ELASTICITY IN 
LB./IN. 2 BETWEEN LOADS 
IN LB./IN. 2 OF 


COMPRES- 
SIVE 

STRENGTH 
lb./in.2 


Cement 


Sand 


Broken 
Stone 


Months 


Days 


100 and 
600 


100 and 
1000 


1000 and 
2000 


1 


2 


4 




7 


1,000,000 






962 


1 


2 


4 


1 




2,083,000 


1,875,000 


1,190,000 


2590 


1 


2 


4 


3 


. . . 


4,167,000 


3,750,000 


2,778,000 


3226 


1 


2 


4 


6 




3,125,000 


3,214,000 


3,571,000 


4847 


1 


3 


6 


1 


. 


2,083,000 


2,143,000 


. 


2364 


1 


3 


6 


3 




3,571,000 


2,812,000 


2,000,000 


2880 


1 


3 


6 


6 




4,167,000 


4,091,000 


1,724,000 


2627 


1 


6 


12 


1 




1,667,000 


1,607,000 




1452 


1 


6 


12 


3 


. 


1,786,000 


1,800,000 




1677 


1 


6 


12 


6 




1,923,000 


1,667,000 




2055 


1 





2 


. 


7 


2,083,000 


1,607,000 


1,087,000 


2638 


1 





2 


1 


. . . 


3,571,000 


3,214,000 


2,778,000 


3600 


1 





2 


3 


. . . 


2,778,000 


2,500,000 


1,724,000 


3800 


1 





2 


6 




3,571,000 


3,461,000 


2,381,000 


4978 


1 


2 


4 


. 


7 


2,500,000 


2,143,000 


1,351,000 


2400 


1 


2 


4 


1 










2933 




2 


4 


3 


. 


3,571,000 


3,214,000 


2,381,000 


3157 


1 


2 


4 


6 


. . . 


4,167,000 


3,750,000 


1,852,000 


3309 


1 


3 


6 


. 


7 


2,273,000 


1,607,000 




1386 


1 


3 


6 


1 


. 


2,273,000 


1,875,000 


1,219,000 


2431 


1 


3 


6 


3 


. . 


2,778,000 


2,812,000 


2,083,000 


2651 


1 


3 


6 


6 




3,125,000 


3,000,000 


1,852,000 


3207 


1 


6 


12 




7 








754 


1 


6 


12 


1 




961,000 


. 


. 


1088 


1 


6 


12 


3 




2,083,000 


1,667,000 




1276 


1 


6 


12 


6 




1,786,000 


1,364,000 


.... 


1097 


1 





2 


. 


10 


2,500,000 


2,368,000 


1,515,000 


3279 


1 





2 


1 


. 


2,273,000 


1,956,000 


1,429,000 


3420 


1 





2 


3 


. 


2,273,000 


2,250,000 


1,515,000 


3144 


1 





2 


6 


. . . 


3,571,000 


3,214,000 


2,273,000 


5360 



* For the method of computing the modulus of elasticity for materials which do not 
conform to Hooke's law see Article 65. 



310 



STRENGTH OF MATERIALS 



The strain diagram of concrete in compression, shown in Fig. 181, 
illustrates the fact that there is no well-defined elastic limit, and that 
the modulus of elasticity changes as the load increases. 

The table on page 309 also illustrates the variation in the modulus 
of elasticity of concrete in compression.* In the first ten tests the 
cement used in making the test specimens was Alpha Portland, in 
the next sixteen it was Germania Portland, and in the remaining 
ones Alsen Portland. 

Problem 301. From the strain diagram of concrete in compression shown in 
Fig. 181, compute the modulus of elasticity at 1800 lb./in. 2 and at 2400 lb./in. 2 
The height of the block tested was 10 in. 

Problem 302. A concrete beam 6 in. x 6 in. in cross section, and with a 68-in. 
span, is supported at both ends and loaded in the middle. The load at failure is 
1008 Ib. Find the maximum fiber stress. 

COMPRESSIVE STRENGTH AND MODULUS OF ELASTICITY OF 
CINDER CONCRETE CUBES 









MODULUS OF ELASTICITY 


COMPOSITION 


AGE 
















COMPRESSIVE 
STRENGTH 


Between 




Cement 


Sand 


Cinders 


Water 


Days 


Ib./in.a 


Loads per 
Square Inch of 


At Highest 
Stress Ob- 














500 and 1000 


served 














Ib. 




1 


2 


4 


n 


38 


1950 


1,786,000 


1,136,000 


1 


2 


4 


n 


38 


2050 


1,923,000 


1,136,000 


1 


2 


4 


n 


224 


2600 


1,471,000 


1,087,000 


1 


2 


4 


H 


224 


2500 


1,563,000 


463,000 


1 


21 


5 


if 


38 


1400 


1,250,000 


. 


1 


21 


5 


if 


38 


1400 


893,000 


. . . 


1 


2| 


5 


ii 


224 


1980 


1,136,000 


893,000 


1 


2 1 


5 


if 


224 


2020 


1,250,000 


694,000 


1 


3 


6 


2 


34 


1200 


781,000 


. 


1 


3 


6 


2 


34 


1330 


1,000,000 


. 


1 


3 


6 


2 


220 


1730 


1,000,000 


694,000 


1 


3 


6 


2 


220 


1560 


735,000 


463,000 



219. Cinder concrete. The preceding table summarizes the results 
of a series of tests made on cinder concrete cubes at the Watertown 



* Watertown Arsenal Report, 1899. 



LIME, CEMENT, AND CONCRETE 



311 



32 


^BliiiiiiiiiffiB 




28 


;||1||||||||| 




24 - 


linM! iujjn iiiin jmj is 






i 




Ijll 


r ~f-~ 






Pi Iffl^ili 






__ X 




8 rSSi-^ 






::::p::::: 


STRAIN DIAGRAM 




4 || 


CONCRETE IN COMPRESSIO 


N 


.002 


-fjqmpr essi |4 n_f n. liip^es I _ I 
.004 .OOG .008 .010 

FIG. 181 


.012 



Arsenal.* The table shows the variation of the modulus of elasticity 
for different stresses. Lehigh Portland cement was used and the 
cubes were set in air. 

220. Concrete building blocks. During the past few years great 
progress has been made in the manufacture and use of concrete 
building blocks. In comparison with stone these have the advan- 
tage of cheapness, ease of manipulation, and beauty of the finished 
product. A type of concrete building block is shown in Fig. 182, 
and illustrates the general characteristics of such blocks. 

* Watertown Arsenal Report, 1903. 



312 



STRENGTH OF MATERIALS 



Few tests have been made on concrete blocks, and but little is 
known as to their durability. The following table is a report of a 
series of tests made at the University of Michigan.* The blocks were 
first tested in flexure, and then an uninjured portion of the block 
was tested in compression. Blocks 3, 4, 5, and 6 were from the same 
mixture, and were composed of one part cement, two parts sand, and 
three parts broken stone. They were all tested after four months. 

TESTS OF CONCRETE BUILDING BLOCKS 



NTMKEK OF 


DISTANCE BE- 
TWEEN SUPPORTS 


STRENGTH IN 
FLEXURE 


STRENGTH 
IN COMPRESSION 


STRENGTH IN 
TENSION 














in. 


Ib. 


lb./in.2 


lb./iii.2 


1 


13 


5450 


604 




2 


24 


3000 


1060 




3 


21 


4100 


705 


121 


4 


21 


4000 


1500 


300 


5 


21 


2900 


940 


237 


6 


21 


3600 


1320 


235 



Problem 303. A concrete building block 24 in. in length and having an effective 
cross section of 8 in. x 10 in. minus 4 in. x 10 in. is tested by being supported at 
both ends and loaded in the middle. The load at failure is found to be 5000 Ib. 
Find the maximum fiber stress, the height of the block being 10 in. 

221. Effect of temperature on the strength of concrete. Concrete 
put in place in cold weather increases in strength very slowly, making 
it necessary to keep the forms on for a much longer time than is 
required when the temperature is 70 or over. The failure of many 
engineers to recognize this fact has been responsible for the early 
removal of forms in cold weather, and in many such cases for a total 
or partial collapse of the structure. An investigation recently car- 
ried out at the Worcester Polytechnic Institute f shows the rate of 
increase of the strength of concrete for temperatures ranging from 
32 F. to 74 F. as summarized in the following table. 



AGE IN DAYS 



TENSILE STRENGTH, lb./in. 2 





32o F. 


36 F. 


47 F. 


74 F. 


1 


608 


679 


897 


1059 


2 


1027 


1232 


1435 


1746 


3 


1127 


1312 


1443 


1491 


4 


1274 


1422 


1587 


1598 


5 


1591 


1662 


1790 


1359 J 


6 


1617 


1689 





15241 



* Concrete, February, 1905. f Eng. News, Vol. LXII, p. 183. \ Probably dried out. 



CHAPTER XIY 

REENFORCED CONCRETE 

222. Object of reenf or cement. The fact that concrete is much 
stronger in compression than in tension has led to attempts to 
increase its tensile strength by imbedding steel .or iron rods in the 
material. This metal reinforcement is so designed as to carry most of 
the tensile stress, and thus plays the same part in a concrete structure 
as the tension members play in a truss. 

It has been found by experiment that reenforced concrete beams 
may be stressed in flexure far beyond the elastic limit* of ordinary 
concrete, and even beyond the stress which would rupture the same 
beam, if not reenforced, without appreciable injury to the material. 
M. Considere, one of the leading French authorities on the subject, 
reports a test of this kind, in which he found that concrete taken 
from the tensile side of a reenforced concrete beam tested in flexure 
was uninjured by the strain. Professor Turneaure, of the University 
of Wisconsin, has found that minute cracks occur on the tension side 
of 'a reenforced concrete beam as soon as the fiber stress reaches the 
point at which non-reenforced concrete would crack, f Experiments 
of this kind seem to indicate that the metal reenforcement carries 
practically all of the tensile stress, as cracks in the concrete must 
certainly reduce its tensile strength to zero at this point. 

223. Corrosion of the metal reenforcement. The maintenance of 
the increased strength of concrete due to the metal reenforcement 
depends upon the preservation of the metal. The corrosion of metal 
imbedded in concrete is thus a matter of the greatest importance in 
connection with reenforced concrete work. It has been found that 
metal thus protected does not corrode even though the concrete be 

* As indicated in Chapter XIII, concrete shows no well-defined elastic limit, i.e. the 
material does not conform to Hooke's law. In this case elastic limit means the arbitrary 
point beyond which the deformations are much more noticeable than formerly. 

t Proc. Amer. Soc.for Testing Materials, 1905. 

313 



314 STRENGTH OF MATERIALS 

subjected to the severest exposure. However, the existence of cracks 
on the tension side of reenforced beams makes the exposure of the 
metal rods possible, and thus adds a new danger to the life of the 
beam ; but the small hairlike cracks that occur after the elastic limit 
of the concrete has been passed probably have no effect in this respect. 
When they become large enough to expose the reenforcement, the 
strength of the beam is endangered. 

224. Adhesion of the concrete to the reenforcement. When a reen- 

forced concrete beam is subjected to stress, there is always a tendency 

to shear horizontally along the reenforcement. This is prevented in 

part by the adhesion between the steel and 

1 -=Sfctz-.^gL. Tzy ii_j^ concrete. Failure sometimes occurs, due to 




this horizontal shear, especially when the 
beam is over-reenforced, i.e. when the area of 
cross section of the reenforcement is large as 
compared with the total area of cross section 
of the beam. When plain round or square rods 
are used, the adhesion between the steel and 
concrete furnishes the only bond. For com- 
FIG. 183 mercial purposes, however, various forms of 

i,Kahn trussed bar; 2, Johnson re g n f orcem ent are ordinarily used to increase 

corrugated bar ; 3, Thaeher 

bulb bar; 4, Ransome twisted tliis bond. Four of these commercial types 
bar are illustrated in Fig. 183. The Johnson, 

Thaeher, and Ransome bars are provided with projections and inden- 
tations to prevent the bar from pulling out of the concrete, while the 
Kahn bar, by means of the projecting arms that extend upward along 
the lines of principal stress in the beam, is also designed to act as 
a truss. Several other commercial types of bar are also in use, but 
all are provided with projections or indentations of some kind to 
prevent slipping. 

Many tests have been made to determine the force necessary to 
pull the various forms of rods from concrete. The following table 
gives the results of pulling-out tests made by Professor Edgar Mar- 
burg, of the University of Pennsylvania.* The rods in this case were 
imbedded centrally in 6 in. X 6 in. concrete prisms 12 in. long, and 
were tested after thirty days. In most cases, except in that of the 

* Proc. Amer. Soc.for Testing Materials, 1904. 



REENFORCED CONCRETE 



315 



plain rods, failure was due to the breaking of the rods or the cracking 
of the concrete. On account of the projections on some of the rods 
these can hardly be called adhesion tests, but should more properly 
be called pulling-out tests. 

As might be expected, the plain rods show the lowest values, since 
any reduction in cross section of the rod, due to the tensile stress 
upon it, largely destroys the adhesion of the concrete. Square reen- 
forcing rods, or those that present sharp angles, are likely to cause 
initial cracks upon the shrinkage of the concrete. To have the 
strongest bond a rod should be round, with rounded projections. 

PULLING-OUT TESTS 









LOAD PER 


REMARKS 






TOTAL LOAD 


LINEAR INCH 


KIND OF 


ROD 


Ib. 


OF ROD 








Ib. 




C 


13,660 


1138 


Elastic limit passed. Concrete 


cracked. 


Johnson . 


J 


12,830 


1069 


Elastic limit passed. Concrete 


cracked. 




L 


9,980 


832 


Concrete cracked. 






{ 


6,280 


524 


Rod pulled out. 




Plain . . 


\ 


6,190 


616 


Rod pulled out. 






I 


5,650 


471 


Rod pulled out. 






f 


10,420 


868 


Rod broke. 




Thacher . 


.<! 


8,890 


741 


Concrete cracked. 






I 


9,970 


831 


Rod broke. 






f 


22,690 


1891 


Concrete cracked. 




Ransome 




16,680 


1390 


Concrete cracked. 






' L 


19,290 


1608 


Rod pulled out. 





225. Area of the metal reinforcement. Since the small hairlike 
cracks mentioned in Article 222 occur early during the flexure of 
a reenforced concrete beam, it is evident that in designing little can 
be allowed for the tensile strength of the concrete. The problem 
becomes one of opposing the compressive strength of the concrete 
and the tensile strength of the reenforcement. This means that 
knowing the safe compressive strength of the concrete and the area 
of the concrete in compression, sufficient steel must be used to carry 
safely a tensile load equal to the compressive load on the concrete. 
Professor Marburg, in the paper referred to in the preceding article, 



316 



STEENGTH OF MATERIALS 



gives 1600 lb./in. 2 for the compressive strength of 6-inch cubes 
thirty days old. A slightly higher value was found for cubes from 
a different mixture. 

From an investigation of the tensile strength of steel reenforcing 
bars, the writer referred to above obtained the following values. 





AREA OF 


ELASTIC 


ULTIMATE 


MODULUS OF 


PERCENTAGE OF 


TYPE OF ROD 


METAL 


LIMIT 


STRENGTH 


ELASTICITY 


ELONGATION 




in. 2 


lb./in.2 


lb./in. 


lb./in.2 


IN 8 IN. 


Plain . . 


.75 


40,500 


60,600 


30,500,000 


23.50 


Johnson 


.54 


65,800 


102,300 


28,500,000 


13.50 


Ransome . 


.76 


58,000 


86,500 


26,000,000 


7.75 


Thacher . 


.59 


31,900 


51,300 


28,500,000 


13.00 



With a 1-3-6 concrete a 1.5 per cent reenforcement of steel, 
having an elastic limit of 33,000 lb./in. 2 , and a 1.0 per cent reenforce- 
ment of steel, having an elastic limit of 55,000 lb./in. 2 , has been 
used without developing the full compressive strength of the concrete.* 
In this case the percentage is figured on the area of concrete above 
the center of the metal reenforcement. This percentage may also be 
figured on the area of cross section of the beam. 

226. Position of the neutral axis in reenforced concrete beams. 
In Article 218 it was pointed out that the modulus of elasticity of 
concrete in compression is not constant. This indicates that in the 
case of flexure the position of the neutral axis changes with the stress, 
at first lying near the center, but moving toward the compression side 
as the load is increased. In a reenforced concrete beam the neutral 
axis also undergoes a displacement, due to the non-homogeneity of 
the cross section, since the moduli of elasticity of steel and concrete 
are not the same. In this case, if the beam is reenforced only on the 
tension side, and the metal reenforcement is designed to carry all the 
tensile stress, the neutral axis usually lies nearer the tension side of 
the beam than the compression side.f 

From tests made at Purdue University, Professor Hatt found the 
ratio of the moduli of elasticity of steel in tension to concrete in 

* Proc. Amer. Soc. for Testing Materials, 1905. 

t See article by S. E. Slocum, entitled " Rational Formulas for the Strength of a Con- 
crete Steel Beam," Engineering News, July 30, 1903. 



KEENFORCED CONCRETE 317 

compression, for certain grades of material to be as follows.* The use 
of this ratio is exemplified in the following article. 

Stone concrete 23 days 8.8 

Stone concrete 00 days 6.6 

Average 7.7 

Gravel concrete 28 days 8.0 

Gravel concrete 90 days 6.2 

Average 7.1 

227. Strength of reenforced concrete beams. Concrete is weak in 
tension and strong in compression, so that when used in the form of 
a beam, the tensile strength controls the strength of the beam. To 
correct for this lack of strength in tension, steel rods are imbedded in 
beams in such a way as to carry the tensile stresses. 

A few years ago engineers believed that the tensile strength of the 
concrete in a beam might be considered in computing the strength of 
the reenforced concrete beam. At present, however, the steel reen- 
forcement is designed to carry all the load in tension and the con- 
crete all the load in compression ; that is, the tensile strength of the 
steel is balanced against the compressive strength of the concrete. 
Since concrete is imperfectly elastic, the stress-strain diagram is not 
quite a straight line in any part of its length. This means that its 
modulus of elasticity is not constant, but changes with the stress. 
The results of many tests show that the stress-strain diagram for 
concrete may be assumed a parabola, so that the compressive stress 
on any section of the beam varies as the ordinates of a parabola. 

On account of the difference in the modulus of elasticity of steel 
and concrete (30,000,000 lb./in. 2 and 2,000,000 lb./in. 2 to 2,500,000 
lb./in. 2 ), the position of the neutral axis changes with the load on the 
beam. In the following analysis the assumptions of the common 
theory of flexure are supposed to hold, with the exception of the 
points stated above. 

Let I = length of span, 

x distance of the neutral axis from the compression face, 
d = effective depth of beam ; that is, the distance from top of 
beam to center of gravity of reenforcement, 

* Jour. Western Soc. Eng., June, 1904. 



318 



STRENGTH OF MATERIALS 



r = ratio of area of steel to that of the effective cross section 

of the beam, 

E s = modulus of elasticity of steel, 
E c = modulus of elasticity of concrete in compression, 
p s = unit stress in metal reinforcement, 
p c = unit compression stress in the concrete at outer fiber, 
e = unit contraction in concrete, and e r unit elongation in the 

steel, 
E c is measured at stress p c . 

The beam is supposed reenforced on the tension side only, and the 
rods are imbedded to a sufficient depth to protect the steel. (This 



p 


- ^ _J_ 


T t 




N.A t _fj 


j _^ X <T <?=*# 


STEEL 






>P o 




,_ _L 




FIG. 184 

depth may be as much as 2^ to 3 in. if the best fire protection is 
desired.) The illustration (Fig. 184) shows the beam supported at 
the ends and loaded in the middle, but the formulas derived apply 
to beams having different loadings and supports. 

Considering the compressive stress to vary as the ordinates of a 
parabola, the total compressive stress in the concrete is C = | p c bx, 
and it may be considered as acting | x (distance of center of gravity 
of the parabolic area) from the top of the beam. The total stress in 
the steel is p s rbd. 

The moment of the stress couple may therefore be written 



-/IZ q- p^XO {(I T iCy, (^) 

or M=p,rdb(d-%x); (b) 

that is, either the compression in the concrete times the distance 
(d | x), or the tension in the steel times the same distance. This 



REENFORCED CONCRETE 



319 



moment of the stress couple must be equal to the moment of the ex- 
ternal forces acting upon either portion of the beam. If the beam is 
supported at the ends and loaded in the middle, and the middle sec- 
tion is considered, then 

2 ,/, 3 \ PI 
-g*)=-> 



or 



PI 
~ 



Equating the summation of horizontal forces (Fig. 184) to zero, 
we have 



or 



But on the assumption that the cross sections of the beam remain 
plane during flexure (see Fig. 185), 

e e' e 

or : 



From definition, 

so that 

and therefore 



x d x 
E. = -& and 




Eliminating p e between (d) and (e), we have 

O 77Y 

x 2 = rd ' (d x). (/) 

3 E c 

Equation (/) may be used to locate the neutral axis in a beam. 
The ratio - is taken by different authorities from 12 to 15. Tests 

made in this country seem to show the lower value as more nearly 
correct, although 15 is usually used. It has also been found that the 
relations of d and x may be expressed approximately by 

x = .52 d. 



320 STRENGTH OF MATERIALS 

Substituting this value in equations (a) and (6), we have 

and M = .80 p 8 rbd?. (h) 

Problem 304. A reenforced concrete beam 8 in. x 10 in. in cross section, and 
15 ft. long, is reenforced on the tension side by six --in. plain steel rounds. The 
steel has a modulus of elasticity of 30,000,000 lb./in. 2 , and the center of the 
reenforcement is placed 2 in. from the bottom of the beam. Assuming that 
E c = 3,000,000 lb./in. 2 , and p c = 600 lb./in. 2 , find from formulas (/) and (a) the 
position of the neutral axis and the moment M. 

NOTE. The moment M corresponds to the moment obtained from the consideration 

of the flexure of homogeneous beams; that is to say, M is the moment of resistance of 
the beam (see Article 44). 

Problem 305. For a stress p c = 2700 lb./in. 2 on the outer fiber of concrete 
in the beam given in Problem 304, find the stress p s in the steel reenforcement. 





P 






\ 


, Pc 1 


T 


T 


^M2 


N.A. 




& T UPcXb 


T 

STEEL J 


/ 


(d-f) 











FIG. 186 

Problem 306. Using the data of Problem 304, locate the neutral axis, and find 
the value of the moment of resistance M under the assumption that the stresses in 
the concrete vary linearly. 

228. Linear variation of stress. It is believed by most engineers 
that it is not necessary to consider that the compressive stress varies 
as the ordinates of a parabola, but that, for working loads, the 
variation is close enough for practical purposes (see Fig. 186). 

Equations (a) and (&) may then be written 



(0 



V) 



BEENFORCED CONCEETE 321 

and equation (/) becomes 

-a* = drjj(d-x). 

In this case it has been found that 

x= ^ d, approximately, 
so that equations (/) and (j) may be written 

\ (ft) 



229. Bond between steel and concrete. The reenforced concrete 
beam should be regarded as a girder. The concrete in compression 
should be regarded as one flange, the steel in tension as the other, 
while the web is made up of concrete. In order that the steel reen- 
forcement may act effectively, it is necessary that there be sufficient 
bond between the steel and concrete to carry the horizontal shear 
occurring along the reenforcernent. The stress that this bond must 
carry is about the same as that carried by the rivets connecting the 
flange and web in a plate girder. 

If y denotes distance along the beam, we know (Article 53) that 

?= 

and so from equations (a) and (m), 

g_&W,_| 

Ay \ 8 

where F is the area of cross section of the reenforcement, or, calling 
d | x, d', this may be written 

F dps = Q 

dy d' 

But F is the rate of change of total stress in the reenforcing bars 

dy 
as y varies along the beam. For unit length of beam, it measures the 

stress transmitted by the concrete to the bars, that is, the bond. 

Let k = number of bars, 
and o = surface of one bar per inch of length. 

Then ok = surface of steel per inch of length of beam, 
and oku = bond, where u is the bond developed per unit area of 
rod surface of bars. 



322 

Then 

For parabolic loading, 

If x = .52 d, 



STEENGTH OF MATEEIALS 

- == oku. 

_Q 

(d | x) ok 



u = 



For the linear variation, 



If x = | d, this becomes 



.8 (okd) 
_Q 

Q 



(A 



l(okd) 

These equations give the unit horizontal shearing stress along the 
reenforcement. From what has been shown previously, this is also 
the unit vertical shearing stress at the reenforcement. 

Turneaure and Maurer * give the following as working stresses 
in concrete beams. 



ULTIMATE 


WORKING 









STRENGTH, 
CONCRETE 


STRESS, 
CONCRETE 


ELASTIC LIMIT, 
STEEL 


SAFE STRESS 
IN STEEL 


SAFE BOND 


COMPRESSION 


COMPRESSION 


lb./in. 


lb./in.2 


lb./in.2 


lb./in.* 


lb./in.2 








2000-2200 


500-600 


32,000 


12,000-15,000 


50-75 for plain ; 
100 for deformed 



The weight of concrete .... 
The weight of reenforced concrete . 



140-150 lb./ft. 3 
150 lb./ft. s 



Problem 307. A concrete beam is 10 x 16 in. in cross section and 20 ft. long. 
It is reenforced with four f-in. steel rods with centers 2 in. above the lower face 
of the beam. The safe compressive strength of the concrete is 600 lb./in. 2 and the 
steel used has an elastic limit of 40,000 lb./in. 2 What single concentrated load 
will the beam carry at its middle ? What tension will be developed in the steel ? 
What shearing stress along the reenforcement ? 

Problem 308. Find what load, uniformly distributed, the beam in the preceding 
problem will carry, and find the tension in the steel and bond for this case. 

230. Strength of T-beams. The T-beam shown in Fig. 187 is much 
used in floor systems in reenforced concrete buildings. Here, as in 

* Principles of Reinforced Concrete Construction, pp. 170-172. 



REENFORCED CONCRETE 



323 



the case of those of rectangular section, the cross sections are assumed 
to remain plane during bending. We have, then, 







e' p s E c d-x 

The tension of the concrete in the web, and the small amount of 
compression when the neutral axis falls below the flange, may be 





7 
















^ 














I 




T 




I I 




/' i 


* C7 


I 









I x 




-/ f ' 








1 




v N.A. 




7 












' 

STEEL 


I 


(. 

1 1 j 7c 7 > 


* 














-*P/ 






-fc^ 















FIG. 187 

neglected without serious error. Then the compression in the flange 
is balanced by the tension in the steel, or C = p s F. The average 

v I c\ 
unit stress in the flange p' c = I x ) and the total compressive 

v / c\ X \ ' 

stress C = ( x 1 ~bc, so that 

x \ 2 / 



Eliminating by means of (r), we have for x, 

PC 

^ _ 2 nFd + cA 
-nF] ' 



where n = and A = area of flange. 

The moment of the internal couple is 

M=cd"=p s Fd". 

The distance d" = d ~x, where x is the distance of the center of 
gravity of the trapezoidal area of stress in the flange from the outer 
fiber. This distance x may be expressed as 



_ c 



3 2x-c 



324 STRENGTH OF MATERIALS 

The resisting moment of the T-beam may now be written 

x) ) (t) 

or M=2i s F(d-x). (u) 

If the neutral axis falls within the flange, then d", the arm of the 

internal couple, will be greater than d , so that a safe approximation 

o 

for the resisting moment is obtained by using 



or 



When the neutral axis falls on the lower edge of the flange, these 
formulas (v) and (w) are exactly true. 

The horizontal shear in the case of T-bearns may be obtained as 
follows. From equation (u) 



_ ~ == ^ I rt d _x) == Q 

dy dy 

Q 

and -T-^ = OKU, 

d x 

where oku has the same meaning as in the case of rectangular 
beams. So that 

Q 



u = 



or, from (w), 



ok (d x) 
Q_ 

~3 



231. Shear at the neutral axis. If the tension in the concrete 
is neglected, in the case of rectangular beams, the horizontal shear 
at the neutral axis must be equal to the horizontal shear along the 
reenforcement. If u r is the unit horizontal shearing stress in the 



REENFORCED CONCRETE 325 

concrete at the neutral axis and b is the width of the beam, then 
bu' = shear per unit length of beam, at the neutral axis, and so 

"'=! 

and therefore for parabolic variation of stress 

Q 



and for linear variation of stress 

o 



In the case of the T-beams, if the tension in the web is neglected, 
the horizontal shear where the web joins the flange must be equal to 
the shear along the steel. Then 



and so 

b' (d - x) 



CHAPTER XV 

BRICK AND BUILDING STONE 

232. Limestone. Limestone is principally a carbonate of lime, made 
up of seashells that have been deposited from water during past 
geological times. Its method of formation has much to do with its 
value as a building material. If it contains no thin layers of clay 
or shale (sedimentary planes), it is likely to be fairly homogeneous in 
structure. But if layers of shale, however small, occur, the material 
is much more quickly weathered. This is especially true if the stone 
be placed at right angles to the position it occupied in the quarry. 

Thin planes of foreign substances are likely to occur in many 
of our best building stones, as may be seen in the rapid deteriora- 
tion of seemingly first-class limestone when used as curbing. Such 
disintegration is caused by a lessening of the adhesion between the 
particles of stone. 

Limestone may be composed of a great percentage of sand cemented 
together by calcareous matter, in which case it is called siliceous lime- 
stone. Under such circumstances chemical action may remove the 
cementing material, thus leaving the stone free to crumble. Marble 
is almost pure limestone. 

Conditions to which a building stone is to be exposed will determine 
the character of the material to be used in any particular structure. 
Rapid freezing and thawing is likely to set up internal strains in the 
material, which may lead to future failure. These strains may be 
caused by unequal expansion or contraction of the particles of the 
stone, or by the freezing and thawing of the water in the stone. The 
formation of ice in the sedimentary planes accounts in a large measure 
for the rapid deterioration of stone. 

Limestone often occurs in very thick layers, as in the case of the 
oolitic limestone found at Bedford, Indiana, where the layers are 
often from 25 to 30 ft. thick. In such cases it is a most valuable 

326 



BRICK AND BUILDING STONE 327 

building stone, especially for bridge piers and other structures where 
large masses of stone are needed. This particular limestone, unlike 
most others, is easily worked, being almost equal to sandstone in 
this respect. 

When limestone is subjected to the atmosphere of a large city, where 
great quantities of coal are used, it is acted upon by the sulphuric 
acid in the air. To determine the effect of this action, a small piece of 
stone, well cleaned, is placed in a 1 per cent solution of sulphuric 
acid and left for several days. If no earthy matter appears, it may be 
concluded that the stone will withstand the action of the atmosphere. 

233. Sandstone. Sandstone consists very largely of grains of sand 
(silicon) cemented together. It has been deposited from water, making 
it homogeneous in structure, and as it occurs in vast beds, it is very 
suitable for building purposes. The ease with which it may be carved 
and worked makes it a much more valuable building material than 
limestone. Various foreign substances, such as iron, manganese, etc., 
give to the stone a variety both in color and texture. Sandstone 
absorbs water much more readily than limestone, and were it not for 
the fact that it occurs in such thick layers, and is therefore almost 
free from sedimentary planes, this might be a serious objection to its 
use. The mean weight of sandstone is 140 lb./ft. 3 ; that of limestone 
is 160 lb./ft. 3 

234. Compression tests of stone. The most common test for a 
building stone is that of subjecting it to a direct crushing force in 
an ordinary testing machine. To prevent local stresses, the specimen, 
which is generally a well-finished cube, is usually bedded in plaster 
of Paris, thin pine boards, or thick paper, and the load at first 
crack and the maximum load are noted. The friction of the bedding 
against the heads of the machine tends to prevent the spreading of 
the specimen near these heads and thus adds to the strength of the 
cube. Great care is necessary in preparing the specimen, in order to 
get the two bearing faces exactly parallel. The stone fractures along 
the 30 line approximately, giving the characteristic fracture of two 
inverted pyramids (Figs. 188 and 189). 

From a series of tests made by Buckley on the building stones of 
Wisconsin,* the average of ten tests on limestone gave an ultimate 

* Buckley, Building Stones of Wisconsin. 



328 



STRENGTH OF MATERIALS 



strength of 23,116 lb./in. 2 , a modulus of elasticity ranging from 
31,500 lb./in. 2 to 1,800,000 lb./in. 2 , and a shearing strength ranging 
from 1735 lb./in. 2 to 2518 lb./in. 2 The average of thirty tests on 
sandstone gave an ultimate strength of 4109 lb./in. 2 , and a modulus 
of elasticity ranging from 32,000 lb./in. 2 to 400,000 lb./in. 2 

From a series of tests on building stone from outside the state of 
Wisconsin, the same report gives the ultimate strength of limestone 
as ranging from 3000 lb./in. 2 to 27,400 lb./in. 2 , and the ultimate 
strength of sandstone from 2400 lb./in. 2 to 29,000 lb./in. 2 This 
report also gives tables showing the effect of freezing and thawing 
on the strength of stone, the effect of sulphuric acid on limestone, 
and the effect of high temperatures on building stone. 

The following table shows the results of a series of compressive 
tests made upon limestone at the Watertown Arsenal.* 



HEIGHT 
in. 


SECTIONAL AKEA 
in .2 


FIRST CRACK 
Ib. 


ULTIMATE STRENGTH 
lb./in. 2 


4.06 


16.4 


361,000 


28,950 


4.08 


16.36 


178,000 


18,496 


3.99 


15.88 


217,200 


13,680 


3.99 


16.04 


219,100 


13,660 


4.01 


15.96 


241,000 


15,320 


4.00 


15.96 


273,400 


17,130 



From another series of tests made at the Watertown Arsenal on 
a different grade of limestone, the average value of the ultimate 
strength was found to be 7647 lb./in. 2 , and the modulus of elasticity 
to be 3,200,000 lb./in. 2 f 

This wide range in the strength of building stone is explained by 
the method of its formation, which makes the character of the stone 
from one locality often differ entirely from that of a neighboring 
locality. Average values of the strength of building stone are there- 
fore of little value, and must be used with a large factor of safety. 

Problem 309. A granite block was tested in compression, the load at first crack 
and at maximum being 263,000 Ib. and 417,400 Ib. respectively. The sectional area 
was 16.4 in. 2 Find the intensity of stress at first crack and at maximum. 



Watertown Arsenal Report, 1900. 



t Watertown Arsenal Report, 1894. 




FIG. 188. Result of Compression 
Test of Limestone 




FIG. 189. Results of Compression Tests of Sandstone 



BRICK AND BUILDING STONE 329 

235. Transverse tests of stone. The use of stone where transverse 
stress is applied calls for some knowledge of its transverse strength. 
A stone may meet the specifications for crushing and yet fail entirely 
when subjected to cross bending, since a beam is in tension on one 
side and in compression on the other. As stone is much stronger in 
compression than in tension, it usually fails in tension under trans- 
verse loading. 

To test the transverse strength of stone, small beams are pre- 
pared usually 1 in. square by 6 or 8 in. long. These are supported 
on knife-edges resting on the platform of the testing machine, and 
the load is applied at the center. Buckley reports limestone beams 
1 in. x 1 in. X 6 in. to have a modulus of rupture of 2000 lb./in. 2 , 
and sandstone beams 1 in. x 1 in. X 4 in. to have a modulus of rupture 
of 1000 lb./in. 2 

236. Abrasion tests of stone. The most extended series of tests 
of stone iii resisting abrasion was made by Bauschinger.* Four-inch 
cubes under a pressure of 4 lb./in. 2 were subjected to the abrasive 
action of a disk having a radius of 19.5 in. and making 200 revo- 
lutions per minute, upon which 20 g. of emery were fed every 10 
revolutions. The loss of volume in cubic inches was as follows. 

Granite 24 dry and .46 wet 

Limestone 1.10 " 1.41 " 

Sandstone . 80 " .64 " 

Brick 38 " .75 " 

Asphalt 60 " 1.62 " 

Abrasion tests of stone have never been standardized, and comparison 
of results of different tests must be made with a full understanding 
of all the conditions affecting the results. 

237. Absorption tests of stone. The absorption test is made to 
determine the amount of water absorbed by the dry stone. In making 
the test the specimen is first heated for several hours at a tempera- 
ture of 212 F., and then placed in water for about thirty hours. The 
increase in the weight of the specimen divided by its weight when 
dry and multiplied by 100 gives the percentage by weight of 'moisture 
absorbed. This percentage for a series of tests varied, for granite, 
from 1.1 to .3 ;.for limestone, from 3.6 to 1.2 ; and for sandstone, from 
13.8 to 1.6. 

* Communications, 1884. 



330 STRENGTH OF MATERIALS 

238. Brick and brickwork. Brick is generally made by tempering 
clay with the proper amount of water, and then molding into the 
desired shape and burning. The tempered clay is used wet, dry, or 
medium, depending upon the kind of brick desired, and these are classi- 
fied as soft mud brick, pressed brick, or stiff mud brick respectively. 
The position of the brick in the kiln may also determine its classifica- 
tion as hard brick, taken from nearest the fire, medium brick from 
the interior of the pile, and soft brick from the exterior of the pile. 

Paving brick is a vitrified clay brick or block somewhat larger 
than the ordinary brick. 

239. Compression tests of brick. For this test a whole or half 
brick is tested edgewise or flat in much the same way as in the 
crushing test for building stone. The faces which are to be in contact 
with the heads of the testing machine are ground perfectly smooth 
and parallel, or are bedded, or both. If plaster of Paris is used, it 
should be placed between sheets of paper to prevent the absorption 
of water by the brick, as this may affect its strength. In any case, in 
testing brick or stone in compression it is desirable to use a spherical 
compression block for one of the heads, so that in case the faces of the 
test piece are not parallel the bearing will adjust itself to bring the 
axis of the test piece into coincidence with the axis of the machine. 
In this case, also, the load at first crack and the maximum load are 
noted. The form of the fractured specimen is also noted ; it is usually 
that of the double inverted pyramid. An imperfect bedding may cause 
the specimen to split vertically into thin pieces. Cardboard cushions 
and soft pine boards are also used in bedding brick for testing. 

The relative value of the kinds of bedding, as indicated by tests 
made at the Watertown Arsenal * on half bricks, may be seen from 
the following table. 

MEAX STRENGTH 

Set in plaster of Paris 5640 lb./in.2 

Set in cardboard cushions 4430 " 

Set in pine wood 4540 " 

The strength of a single brick in compression cannot be taken as 
a criterion of its strength in an actual structure, since its strength in 
that case must depend somewhat upon the mortar used. If the 
mortar is soft and flows (i.e. is squeezed out), the brick may fail in 

* Watertown Arsenal Report, 1901. 



BEICK AND BXJILDIKG STOKE 



331 



tension, due to the lateral flow of mortar, instead of in compression. 
From a series of thirty-eight tests made at the Watertown Arsenal* 
on piers of common brick, it was found that the maximum compres- 
sive strength varied from 964 lb./in. 2 to 2978 lb./in. 2 The mortar 
in this case was composed of one part Rosendale cement and two 
parts sand. The bricks used in these piers developed only one half 
their compressive strength. The compressive strength of soft brick 
may go as low as 500 lb./in. 2 , and that of paving brick as high as 
15,000 lb/in. 2 , when used in piers. 

The following table gives the results of tests of the compressive 
strength of common brick made at the Watertown Arsenal. The 
compressed surfaces were bedded in plaster of Paris, and the bricks 
were tested whole. 



COMPRESSIVE STRENGTH OF COMMON BRICK 



NUMBER 

OF 

BRICK 


DIMENSIONS 


SECTIONAL 
AREA 
in.z 


LOAD AT 
FIRST CRACK 
Ib. 


ULTIMATE STRENGTH 


Height 
in. 


Compressed Sur- 
face 


Total 
Ib. 


Ib./in.z 


in. 


in. 


1 


2.50 


4.22 


8.43 


35.57 


86,000 


186,900 


5,250 


7 


2.48 


4.12 


8.57 


35.31 


107,000 


257,200 


7,280 


13 


2.33 


3.99 


8.47 


33.80 


269,000 


309,500 


9,150 


19 


2.27 


4.04 


8.19 


33.09 


442,000 


609,000 


18,400 


22 


2.30 


4.02 


8.19 


32.92 


93,000 


446,100 


13,550 


25 


2.43 


4.11 


8.67 


35.63 


108,000 


234,800 


6,590 


28 


2.32 


4.09 


8.36 


34.19 


191,000 


361,500 


10,570 


31 


2.55 


4.02 


8.51 


34.21 


216,000 


312,000 


9,120 


34 


2.41 


4.18 


8.48 


35.45 


143,000 


181,000 


5,110 


37 


2.38 


4.00 


8.33 


33.32 


164,000 


282,500 


8,480 


43 


2.46 


4.12 


8.57 


35.31 


63,000 


224,500 


6,360 


45 


2.41 


4.14 


8.57 


35.48 


96,000 


242,700 


6,840 


48 


2.48 


4.15 


8.59 


35.65 


175,000 


229,900 


6,450 


52 


2.36 


4.08 


8.50 


34.68 


142,000 


207,000 


5,970 


54 


2.60 


4.05 


8.49 


34.38 


144,000 


175,700 


5,110 


57 


2.50 


4.16 


9.04 


37.61 


282,000 


653,000 


17,360 


60 


2.45 


4.25 


8.92 


37.91 


138,000 


686,000 


18,100 


*63 


2.49 


4.07 


8.70 


35.41 


185,000 


216,100 


6,100 


09 


2.57 


4.10 


8.50 


34.85 


163,000 


180,600 


5,180 


75 


2.58 


4.14 


8.58 


35.52 


190,000 


259,900 


7,320 


81 


2.37 


4.20 


8.54 


35.87 


148,000 


219,800 


6,130 



Watertown Arsenal Report, 1884. 



332 



STRENGTH OF MATERIALS 



The compressive strength here ranged from 5000 lb./in. 2 to 18,000 
lb./in. 2 Average values for the strength of different kinds of brick 
in compression might be given as follows : soft brick, 900 lb./in.' 2 ; 
hard brick, 3250 lb./in. 2 ; and vitrified brick, 17,500 lb./in. 2 The 
latter includes paving brick. 

Problem 310. The following bricks were tested in compression. 

(a) Red face brick: sectional area, 28.45 in.' 2 ; load at first crack, 379,000 Ib. ; 

load at maximum, 384,600 Ib. 
(6) Vitrified brick : sectional area, 27.46 in. 2 ; load at first crack, 72,000 Ib. ; load 

at maximum, 230,000 Ib. 
(c) Paving brick : sectional area, 26.72 in. 2 ; load at first crack, 51,000 Ib. ; load 

at maximum, 148,000 Ib. 
Find the intensity of stress at first crack and at maximum load in each case. 

240. Modulus of elasticity of brick. As in the case of stone and 
concrete, the modulus of elasticity of brick in compression is not 
constant, but varies to some extent with the load. On account of 
this variation it is hard to give average values for the modulus of 
elasticity of brick, especially as the materials and methods of manu- 
facture are so varied. Therefore in stating the modulus of elasticity 
it is also necessary to state the corresponding load. Strictly speaking, 
brick, stone, and concrete have no modulus of elasticity. 

The table below is the result of a series of tests of dry-pressed and 
mud brick, tested edgewise in compression, and gives the modulus of 
elasticity for loads between 1000 lb./in. 2 and 3000 lb./in. 2 , and also 
at the highest stress observed. 

MODULUS OF ELASTICITY FOR BRICK 









MODULUS OF ELASTICITY 








WEIGHT 




COMPRES- 








lb./m. 2 




KIND OF BRICK 


POSITION IN 
KILN 


PER CUBIC 
FOOT 




SIVE 
STRENGTH 


Between Loads 


At Highest 






Ib. 


of 1000 and 


Stress 


lb./in. 2 








3000 lb./in. 2 


Observed 




Dry pressed 


Top . . . 


128.3 


3,125,000 


3,271,000 


10,300 


(i 


^ down . 


127.2 


3,125,000 


2,846,000 


8,740 


it 


| down . . 


124.3 


2,222,000 


2,174,000 


5,940 


it 


Bottom . . 


119.8 


1,205,000 




3,480 


Mud . . . 


Top . . . 


144.3 


10,000,000 


8,654,000 


19,170 


u 


\ down . 


136.4 


7,692,000 


7,576,000 


15,670 


u 


| down . 


130.6 


5,263,000 


4,545,000 


10,420 


It 


Bottom . .. 


125.4 


4,545,000 


3,977,000 


10,870 



BRICK AND BUILDING STONE 



333 



Problem 311. A dry-pressed brick of sectional area 9.72 sq. in. was tested in 
compression endwise. Measurements were taken on a gauged length of 5 in. and 
the following data obtained. 



APPLIED LOADS 


IN GAUGED LENGTH 


REMARKS 


Total 
Ib. 


lb./in.* 


Compression 
in. 


Set 
in. 


972 


100 


0. 


0. 


Initial load 


1,944 


200 


.0003 


0. 




3,888 


400 


.0007 


0. 




5,832 


600 


.0012 


0. 




7,77G 


800 


.0015 


0. 




9,720 


1,000 


.0017 


0. 




11,064 


1,200 


.0020 






13,608 


1,400 


.0024 







15,552 


1,600 


.0028 







17,496 


1,800 


.0030 






19,440 


2,000 


.0033 


0. 




21,384 


2,200 


.0036 






23,328 


2,400 


.0039 







25,272 


2,600 


.0042 







27,216 


2,800 


.0045 






29,160 


3,000 


.0049 


0. 


E (1000- 3000)= 3,125,000 Ib. /in.* 


31,104 


3,200 


.0051 






33,048 


3,400 


.0054 






34,992 


3,000 


.0057 







36,936 


3,800 


.0060 






38,880 


4,000 


.0063 


0. 




40,824 


4,200 


.0060 


.... 




42,768 


4,400 


.0069 






44,712 


4-,600 


.0072 






46,656 


4,800 


.0075 






48,600 


5,000 


.0078 


.0001 




50,544 


5,200 


.0081 







52,488 


5,400 


.0084 






54,432 


5,000 


.0087 






56,376 


5,800 


.0090 







58,320 


0,000 


.0093 


.0001 




60,264 


6,200 


.0097 






62,208 


6,400 


.0100 






64,152 


6,600 


.0104 







66,096 


6,800 


.0107 







68,040 


7,000 


.0110 


.0002 




69,984 


7,200 


.0113 






71,928 


7,400 


.0117 







73,872 


7,600 


.0120 






75,816 


7,800 


.0124 






77,700 


8,000 


.0127 


.0003 


(1000-8000) = 3,271,000 Ib. /in.* 


100,100 


10,300 






Ultimate strength 







Draw the strain diagram. Compute the modulus of elasticity between loads of 
1000 lb./in.2 and 3000 lb./in. 2 , and also between 6000 lb./in. 2 and 8000 lb./in.a 



334 STRENGTH OF MATERIALS 

241. Transverse tests of brick. Bricks are tested transversely 
by supporting them edgewise or flatwise upon two knife-edges and 
applying the load centrally by means of an ordinary testing machine. 
Care must be taken to provide suitable bearing surfaces for the knife- 
edges, in order to prevent local failure. In this test the upper fibers 
are in compression and the lower fibers in tension, and since brick is 
stronger in compression than in tension, failure is caused by rupture 
of the tension face. The fiber stress is computed from the formula 

Pie 

P = J> 

where P is the breaking load in pounds, I is the length of span in 
inches, e is half the height, and I is the moment of inertia of a cross 
section. The fiber stress on the outer fiber at failure is usually called 
the modulus of rupture. 

For paving brick the modulus of rupture varies from 1000 lb./in. 2 
to 3000 lb./in. 2 For pressed brick, common brick, and medium brick 
the modulus of rupture varies from 300 lb./in. 2 to 1200 lb./in. 2 

The shearing strength of various grades of brick varies from 
300 lb./in. 2 to 2000 lb./in. 2 

Problem 312. A brick having a depth of 2.23 in. and a breadth of 3.95 in. was 
loaded centrally on a span of 6 in. The ultimate load was 1645 Ib. Find the 
modulus of rupture. 

242. Rattler test of brick. Paving bricks were formerly tested in 
abrasion in order to determine their ability to withstand wear. This 
test, however, does not approach the conditions of actual service, 
which consist of the impact of horses' feet as well as the abrasive 
action of traffic. To meet these conditions the rattler test was devised. 
The testing machine consists of a cast-iron barrel mounted horizon- 
tally, and the test is made by placing the brick, together with some 
harder material, such as cast iron, in the machine and revolving 
it at a certain speed for a certain length of time. The ratio of the 
amount of material broken or worn off in this way to the original 
weight of the brick put into the machine indicates the value of the 
brick in withstanding the conditions of service. 

The charge usually consists of nine paving bricks or twelve other 
bricks, together with 300 Ib. of cast-iron blocks, the volume of the 



BRICK AND BUILDING STONE 335 

bricks being equal to about 8 per cent of the volume of the machine. 
The cast-iron blocks are of two sizes, the larger being about 2J- in. 
square and 4J- in. long, with rounded edges and weighing at first 
7J- Ib. The smaller are about IJ-in. cubes, with rounded edges. About 
225 Ib. of the smaller size and 75 Ib. of the larger size are used ; 
1800 revolutions are required, and must be made at the rate of about 
30 per minute.* 

During the first 600 revolutions the effect of the rattler action on 
the brick is to chip off the corners and edges. Thereafter the action 
is more nearly abrasive. 

243. Absorption test of brick. A brick which absorbs a great 
amount of water is likely to be weakened and injured by frost. To 
measure the amount of absorption, a dry brick is taken and a deter- 
mination of its absorbing capacity made, as in the case of stone 
(Article 237). 

Ordinary brick will absorb from 10 to 20 per cent of its own 
weight, and paving brick from 2 to 3 per cent. 

This test is now little used, since a brick that fails in the absorp- 
tion test is of such poor quality that it will also fail when subjected 
to the crushing and cross-bending tests. 

* See specifications of the National Brick Manufacturers' Association for rattler test. 



CHAPTER XVI 

TIMBER 

244. Structure of timber. An examination of the cross section of 
a tree usually shows that it is made up of a rather dark interior core, 
or heartwood, and a lighter exterior portion, or sapwood, surrounded by 
the bark. In some species, such as the oaks, radial lines, called 
medullary rays, are seen running from the center toward the bark. 
If the cross section happens to be near a knot or other defect, this 
normal structure may be changed. If, however, no knots are present, 
a closer examination shows that both the sapwood and heartwood are 
made up of concentric rings, called annual rings, and that this appear- 
ance is due to a difference in structure. Part of the ring is seen to 
be denser than the rest, and, in fact, it is this difference in density 
which gives the section its characteristic appearance. 

The annual rings in one stick of a certain species may be more 
widely separated than those in another stick of the same species, and 
the relative thickness of heartwood and sapwood may differ in different 
sticks. This indicates that the structure of timber varies considerably, 
and that therefore the physical properties also vary. This wide varia- 
tion is seen in all substances found in nature, one instance of which 
has been shown in the case of natural stone. An investigation of the 
physical properties of such substances, therefore, is more difficult than 
that of a more homogeneous substance. However, the extensive use 
of timber as a structural material makes a knowledge of its structure 
and properties of the utmost importance. 

245. Annual rings. Each of the concentric rings in timber repre- 
sents the growth of one year. The inner or less dense portion repre- 
sents the more rapid spring growth, while the outer dense portion 
represents the slower summer and fall growth. The number of rings 
per inch indicates the rate of growth for that number of years. If 
the number of rings per inch be few, the growth has been rapid and 

336 



TIMBER 337 

the spring growth predominates, making the wood somewhat weak. 
If, on the contrary, the number of rings per inch be many, a slow 
growth is indicated and there is a greater amount of the dense, strong 
summer and fall wood. The number and character of the annual 
rings may thus give some idea of the strength of a piece of timber. 

246. Heartwood and sap wood. The heartwood of a tree may be 
considered a lifeless conical core, which is increased each year by the 
addition of a portion of the outer sapwood. Both the sapwood and 
heartwood contain small tubes that extend from the roots of the tree 
to the branches. These tubes in the sapwood carry water charged 
with nourishment to the branches and growing parts of the tree. In 
the heartwood the tubes no longer act as conveyors, although they 
still contain moisture. The heartwood is the mature wood and is 
more valuable for structural purposes. 

247. Effect of moisture. It is well known that green wood is not 
as strong as the same wood when seasoned, which indicates that the 
effect of moisture in timber is to lessen its strength. A live tree as 
it stands in the forest contains a great deal of moisture. When it 
has been cut, sawed, and dried, most of this moisture has evaporated, 
but considerable still remains, and however well seasoned timber may 
be, it will still contain some moisture. 

In making tests of timber, therefore, it is necessary to determine 
the percentage of moisture in order that the results- may be compared 
with the results of other tests. This is determined by cutting a 
small piece from the uninjured portion of the test piece and weighing 
before and after thorough drying. The difference in weight divided 
by the dry weight and multiplied by 100 gives the percentage of 
moisture. 

248. Strength of timber. The strength of timber depends upon the 
amount of heartwood or sapwood, knots (sound or loose), wind shakes 
and checks, cracks, or any defect that breaks the continuity of the 
fiber. In general, the strength of timber is indicated by its weight, 
the heaviest timbers being the strongest. Timber is strongest along the 
grain both in tension and compression, as will be seen in what follows. 

It has been found that values obtained for the strength of timber 
by testing small, carefully selected test pieces are much higher than 
those obtained by testing large commercial timbers. This is what 



338 



STRENGTH OF MATERIALS 



might be expected, since the larger commercial pieces contain knots 
and other defects not found in the selected test pieces. It has been 
found also that the place and conditions of growth, time of felling, 
method and time of seasoning, and many other factors have each 
some effect upon the strength of timber. Since the weight of timber 
is an indication of its strength, some idea of the relative strength 
of the more common species may be obtained by referring to the 
table given below.* 

WEIGHT OF KILN-DRIED WOOD OF DIFFERENT SPECIES 





APPROXIMATE 


Specific 
Weight 


Weight of 


1 cu. 
ft. Ib. 


1000 ft. of 
Lumber 
Ib. 


(a) Very heavy woods : 








Hickory, oak, persimmon, osage orange, black 








locust, hackberry, blue beech, best of elm, 








ash 


0.70-0.80 


42-48 


3700 


(6) Heavy woods : 








Ash, elm, cherry, birch, maple, beech, walnut, 








sour gum, coffee tree, honey locust, best of 








Southern pine, tamarack . 


0.60-0.70 


36-42 


3200 


(c) Woods of medium weight : 








Southern pine, pitch pine, tamarack, Douglas 








spruce, western hemlock, sweet gum, soft 








maple, sycamore, sassafras, mulberry, light 








grades of birch and cherry 


0.50-0.60 


30-36 


2700 


(d) Light woods : 








Norway and bull pine, red cedar, cypress, hem- 








lock, the heavier spruce and fir, redwood, 








basswood, chestnut, butternut, tulip, catalpa, 








buckeye, heavier grades of poplar .... 


0.40-0.50 


24-30 


2200 


(e) Very light woods : 








White pine, spruce, fir, white cedar, poplar . 


0.30-0.40 


18-24 


1800 



249. Compression tests. Compression tests are made on short blocks 
and long columns. For the short-block test the piece is placed in an 
ordinary testing machine between the moving head and the platform, 



*Bureau of Forestry, Bulletin No. 1 D, "Timber.' 



TIMBER 



339 



with its ends as nearly parallel as possible, and the compression is 
measured by an ordinary compressornet6T, or similar instrument for 
measuring the lowering of the moving head. To provide for the non- 
parallelism of the ends, it is well to use a spherical bearing for one 
of the bearing ends. This will insure the proper " lining up " of the 
specimen so that the compression will be along the grain. 

A strain diagram may be drawn by plotting loads in lb./in. 2 as 
ordinates and the corresponding relative compressions as abscissas. 
The elastic limit, modulus of elasticity, modulus of resilience, and 
maximum strength may then 
be obtained from the diagram 
in the usual manner. Failure is 
either due to a splitting of the 
specimen or to a shearing off at 
an angle of about 30 to the 
horizontal (Fig. 190). The latter 
is the characteristic failure for 
green timber. 

The tests on long columns are 
made in much the same way as 

the tests on short blocks. Provision is made for fixing the ends of 
the columns so as to give the standard end conditions, namely, square 
ends, round ends, pin and square ends, etc. In either case sufficient 
data is taken to get a load-deflection curve by measuring the deflec- 
tions at the center corresponding to selected load increments. These 
deflections are usually measured in two directions at right angles to 
each other.* 

Problem 313. Fig. 191 represents the results of compression tests of pine, 
poplar, and oak, plotted with loads in pounds as ordinates and compression in 
inches as abscissas. The blocks were all 7 in. high, with an area of cross section as 
follows : pine, 2 in. x 1.48 in. ; poplar, 2 in. x 1.48 in. ; oak, 2 in. x 1.47 in. Redraw 
the curves, plotting the loads in lb./in. 2 as ordinates and the corresponding unit 
compressions in inches as abscissas. Determine for each material the elastic limit, 
the modulus of elasticity, and the modulus of elastic resilience. Also compare the 
results obtained with the results reported for these materials in compression in 
the following tables. 





FIG. 190 



* For a report of the tests that have been made on full-sized timber columns the 
student is referred to Lanza's Applied Mechanics. 



340 STRENGTH OF MATERIALS 

250. Flexure tests. Flexure tests are usually made by supporting 
a rectangular piece at both ends and loading it in the middle, care 
being taken to guard against local failure at the supports and at the 
point of application of the load. This local failure may be prevented 
by inserting some kind of metal plate between the beam and the 
knife-edge. The deflections of the beam for specified loads are meas- 
ured by means of a deflectorneter, usually measuring to .01 in. or 
.001 in. From the data obtained from a test, a strain diagram may 
be drawn by plotting loads in pounds as ordinates and deflections in 
inches as abscissas. The fiber stress for any load within the elastic 
limit is determined, for central loading, from the formula 

Pie 
P = I> 

and the modulus of elasticity from the formula (Article 67) 

,**-. 

4:8 DI 

The formulas used to determine the fiber stress in the case of the 
flexure of beams ( = J/ max ) are true only within the elastic limit 



e 

of the material. They are used, however, to determine the fiber stress 
beyond the elastic limit, although they are only approximately true 
beyond this limit. The value of the fiber stress at rupture as deter- 
mined by the formula is usually designated as the modulus of rupture 
(Article 65) ; it is expressed in lb./in. 2 

On account of the peculiar structure of timber the character of 
the fracture due to a failure in flexure is rather difficult to predict. 
In case the specimen is free from knots and the grain is parallel to the 
length of the piece, failure from concentrated central loading is likely 
to take place either on the tension or the compression side, or both. 
It may happen, however, even in the case of such a perfect specimen as 
indicated, that failure will be due to horizontal shear. In such cases 
shearing takes place along the spring growth of one of the annual 
rings. This may have been weakened previously by wind shakes. 

If part of the beam is sapwood and part heartwood, the fracture 
will be influenced thereby, due to the difference in the strength of 
the two portions. A cross grain may cause a failure due to splitting. 



TIMBER 



341 







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"r& COMPRESSION TEST o 


i DIAGRAM 
p PINE, POPLAR, AND OAK 


iiiiiiiiiiiiiiiiiBi 


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12345 
FIG. 191 



342 



STRENGTH OF MATERIALS 




Knots of any kind near the central portion of the beam may determine 
the fracture and cause the beam to break off almost squarely. No 
law has yet been determined which will give the effect of knots of 

various sizes on the strength 
of timber. 

Some characteristic failures 
in flexure are shown in Fig. 
192. The lower beam shows a 
normal failure on the tension 
side. The two upper beams 
show the fracture of a some- 
what more brittle material, the 
fracture being influenced by 
the presence of knots. The 
upper beam also shows a com- 
pression failure. 



FIG. 192 Problem 314. A rectangular 

pine beam, width 1.48 in., height 

1.99 in., and span 30 in., was tested in flexure by being supported at both ends and 
loaded in the middle, and the following data obtained. Draw the strain diagram, 
plotting loads in pounds as ordinates and deflection in inches as abscissas. Locate 
the elastic limit and compute the fiber stress on the outer fiber at the elastic limit. 
Also compute the modulus of rapture, the modulus of elasticity, and the modulus 
of elastic resilience. 



CENTRAL LOAD 
Ib. 


DEFLECTION AT 
CENTER 

in. 


CENTRAL LOAD 
Ib. 


DEFLECTION AT 
CENTER 
in. 


100 


.034 


900 


.305 


200 


.061 


1000 


.341 


300 


.097 


1100 


.393 


400 


.132 


1200 


.451 


500 


.166 


1300 


.583 


600 


.201 


1400 


.670 


700 


.234 


1500 


.810 


800 


.270 


1585 (maximum) 


.952 



251. Shearing tests. The shearing strength of timber parallel to 
the grain is usually measured by finding the force necessary to cause 
a small projecting block of the material to shear off along the grain. 




T1MBEB, 343 

In this case the line of action of the force is parallel to the grain. 
The intensity of stress is obtained by dividing the force by the area 
of the sheared surface. 

252. Indentation tests. Indentation tests are intended to show 
the crushing strength of timber perpendicular to the grain. A rec- 
tangular piece of the tim- 
ber is usually chosen, 

and a metal block whose 

width equals the width 

of the specimen is pressed 

into it by an ordinary 

testing machine. Con- FIG 193 

venient load increments 

are taken, and these, together with the corresponding compressions, 

give sufficient data for a load compression curve from which the 

elastic properties may be determined. Fig. 193 illustrates a specimen 

that has been tested in compression perpendicular to the grain. 

253. Tension tests. Tension tests of timber are seldom made on 
account of the difficulty of obtaining satisfactory test pieces. The 
specimens to be tested must be much larger at the ends than in the 
middle in order to provide for attachment in the heads of the testing 
machine, and for this reason the piece is likely to fail by the shearing 
off of the enlarged ends, or by the pulling out of the fastenings. 
This test, therefore, is little used, the flexure test being relied upon 
to furnish information regarding the tensile strength of timber. 

254. European tests of timber. As early as the middle of the 
eighteenth century tests to determine the strength of timber were 
made in France. This work was done for the most part from a scien- 
tific standpoint. The most important European tests were carried out 
by Bauschinger in his laboratory at Munich, from 1883 to 188 7. The 
object of these tests was to determine the effect of the time of felling 
and conditions of growth upon the strength of Scotch pine and spruce. 
From these tests Bauschinger drew the following conclusions. 

1. Stems of spruce or pine which are of the same age at equal diameters, 
and in which the rate of growth is about equal, have the same mechanical 
properties (when reduced to the same moisture contents) , irrespective of local 
conditions of growth. 



344 



STRENGTH OF MATERIALS 



2. Stems of spruce or pine which are felled in winter have, when tested 
two or three months after the felling, about 25 per cent greater strength than 
those felled in summer, other conditions being the same. 

He notes, however, that later tests may change these conclusions 
somewhat. 



AVERAGE RESULTS OF TIMBER TESTS MADE FOR THE 
TENTH CENSUS 



NAME OF SPECIES 


TRANSVERSE TESTS 


COMPRESSION TESTS 


Modulus of 
Rupture 
Ib./m. 2 


Modulus of 
Elasticity 
lb./in. 2 


Compression 
Parallel to 
Grain 

lb./in. 2 


Compression 
Perpendicu- 
lar to Grain 
lb./in. 2 


Poplar .... 


9,400 
8,340 
7,540 
16,500 
14,640 
7,580 
9,330 
12,200 
10,000 
11,900 
12,450 
7,000 
16,600 
11,770 
15,100 
14,900 
8,100 
11,100 
12,250 
15,450 
9,480 
13,270 
13,150 
11,800 
10,440 
16,200 


1,330,000 
1,172,000 
1,158,000 
2,250,000 
1,800,000 
873,000 
1,300,000 
1,262,000 
1,200,000 
1,560,000 
1,445,000 
790,000 
1,912,000 
1,300,000 
1,640,000 
1,450,000 
1,225,000 
1,400,000 
1,567,000 
1,901,000 
1,138,000 
1,870,000 
1,917,000 
938,000 
1,450,000 
1,730,000 


5000 
5190 
5275 
8800 
6850 
4580 
6630 
6780 
7540 
8000 
7670 
6400 
8360 
7100 
6800 
6960 
4800 
5580 
6100 
8300 
5400 
7780 
7400 
6300 
5000 
6770 


1120 
880 
2000 
3600 
2580 
1580 
1880 
2780 
2250 
2680 
2330 
2700 
3500 
3100 
2700 
2980 
1050 
1450 
1520 
1920 
1100 
1750 
1480 
2000 
. 1100 
2840 


Basswood 
Ironwood 


Su^ar maple 


White maple .... 
Box elder 


Sweet sum 


Sour gum .... 


White ash 


Black walnut .... 
Slippery elm .... 
Sycamore 
Hickory (shellbark) . 
White oak 


Red oak 


Black oak 


"White pine 


Yellow pine 


Loblolly pine .... 
Long-leaved pine . . 
Hemlock . 


Red fir 


Tamarack 


Red cedar 


Cottonwood 


Beech 




Averages of all species 
given above . . . 


11,800 


1,445,000 


6600 


2110 



TIMBER 345 

255. Tests made for the tenth census. In the United States, tests 
were made for the tenth census on four hundred and twelve species 
of timber. The test specimens were all small, selected pieces, 1.57 in. 
X 1.57 in. in cross section, and 43 in. long, and were seasoned in a 

dry, cool building for two years. On account of the number of 
species tested the results obtained are not conclusive, but should 
be taken as indicating the probable values for the strength of the 
timbers tested. On page 344 is given a table of the averages for some 
of the species tested. Since the test pieces were all small, selected 
specimens, the results are probably higher than would have been 
obtained from larger commercial specimens. 

In the transverse tests the specimens were supported at both ends 
and loaded in the middle, the span being 39.37 in. The compression 
tests parallel to the gram were made on pieces 1.57 in. x 1.57 in. in 
cross section, and 12.6 in. long. Indentation tests were made on 
pieces 1.57 hi. x 1.57 in. in cross section and 6.3 in. long. The test 
pieces in the latter case rested upon the platform of the testing 
machine, and the tests were made by crushing perpendicular to the 
grain with a plate 1.57 in. x 1.57 in. in size, by lowering the moving 
head of the machine. 

256. Tests made by the Bureau of Forestry. The most extensive 
series of timber tests that has ever been undertaken has been begun 
by the United States Department of Agriculture under the direction 
of the Bureau of Forestry. These tests were begun in 1891, under 
the direction of Professor J. B. Johnson, at St. Louis. Thirty-two 
species were tested and 45,000 tests were made. The material was 
selected with special reference to the conditions under which the 
trees were grown, and the test pieces were small, selected speci- 
mens. The table on page 346 gives the average results of some of 
the tests.* 

In the table the results have been reduced to an amount of moist- 
ure equivalent to 12 per cent of the dry weight. 

A comparison of this table with that of the tenth census shows as 
close an Agreement in most cases as might be reasonably expected 
when the variability of timber is considered, and serves to extend 
and verify the results of the previous work. 

* U. S. Forestry Circular, No. 15. 



346 



STRENGTH OF MATERIALS 



RESULTS OF TIMBER TESTS MADE BY THE UNITED STATES 
BUREAU OF FORESTRY 





TRANSVERSE TESTS 


COMPRESSION 


SHEARING 






TESTS 


TESTS 


SPECIES 


Modulus of 


Modulus of 


Compres- 
sion 


Compres- 
sion Per- 


Shearing 




Rupture 


Elasticity 


Parallel to 


pendicular 


along the 

r\ :,, 




Ib./in." 


Ib./iii.* 


Grain 


to Grain 


vi i am 








Ib./iii. 2 


lb./in.2 


lb./in. 2 


Long-leaf pine . . 


12,600 


2,070,000 


8,000 


1260 


835 


Cuban pine . . . 


13,600 


2,370,000 


8,700 


1200 


770 


Short-leaf pine . . 


10,100 


1,680,000 


6,500 


1050 


770 


Loblolly pine . . . 11,300 


2,050,000 


7,400 


1150 


800 


White pine ... 7,900 


1,390,000 


5,400 


700 


400 


Red pine .... 9,100 


1,620,000 


6,700 


1000 


500 


Spruce pine . . . 10,000 


1,640,000 


7,300 


1200 


800 


Bald cypress . . . 7,900 


1,290,000 


6,000 


800 


500 


White cedar . . . 6,300 


910,000 


5,200 


700 


400 


Douglas spruce . . 7,900 


1,680,000 


5,700 


800 


500 


White oak ... 13,100 


2,090,000 


8,500 


2200 


1000 


Overcup oak . . . 11,300 


1,620,000 


7,300 


1900 


1000 


Post oak .... 12,300 


2,030,000 


7,100 


3000 


1100 


Cow oak .... 11,500 


1,610,000 


7,400 


1900 


900 


Red oak .... 


11,400 


1,970,000 


7,200 


2300 


1100 


Texan oak 


13,100 


1,860,000 


8,100 


2000 


900 


Yellow oak . . . 


10,800 


1,740,000 


7,300 


1800 


1100 


Water oak 


12,400 


2,000,000 


7,800 


2000 


1100 


Willow oak . 


10,400 


1,750,000 


7,200 


1600 


900 


Spanish oak . . . 


12,000 


1,930,000 


7,700 


1800 


900 


Shagbark hickory . 


16,000 


2,390,000 


9,500 


2700 


1100 


Mockernut hickory . 


15,200 


2,320,000 


10,100 


3100 


1100 


Water hickory . 


12,500 


2,080,000 


8,400 


2400 


1000 


Bitternut hickory . 


15,000 


2,280,000 


9,600 


2200 


1000 


Nutmeg hickory . 


12,500 


1,940,000 


8,800 


2700 


1100 


Pecan hickory . . 


15,300 


2,530,000 


9,100 


2800 


1200 


Pignut hickory . 


18,700 


2,730,000 


10,900 


3200 


1200 


White elm .... 


10,300 


1,540,000 


6,500 


1200 


800 


Cedar elm .... 


13,500 


1,700,000 


8,000 


2100 


1300 


White ash . . 


10,800 


1,640,000 


7,200 


1900 


1100 


Green ash .... 


11,600 


2,050,000 


8,000 


1700 


1000 


Sweet gum . . . 


9,500 


1,700,000 


7,100 


1400 


800 



The effect of the presence of moisture on the strength of timber 
was also investigated when these tests were made by testing some of 



TIMBER 



347 



the foregoing species endwise in compression while green. The fol- 
lowing table gives the results of these tests in lb./in. 2 The pieces 
contained over 40 per cent of moisture. A comparison of the results 
obtained from these tests with those reported in the preceding table 
shows that the compressive strength has been diminished from 50 
to 75 per cent by the presence of the given percentage of moisture. 



COMPRESSIVE TESTS OF GREEN TIMBER 



SPECIES 


NUMBER OF 
TESTS 


HIGHEST 
SINGLE 
TEST 
Ib./in.z 


LOWEST 
SINGLE 
TEST 
Ib./in.a 


AVERAGE 

OF ALL 

TESTS 
Ib./in.a 


Long-leaf pine 
Cuban pine 


86 
38 


7300" 
6100 


2800 
3500 


4300 

4800 


Short-leaf pine 
Loblolly pine 
Spruce pine 


8 
69 
71 


4000 
6500 
4700 


3000 
2600 
2800 


3300 
4100 
3900 


Bald cypress 


280 


8200 


1800 


4200 


White cedar . 


34 


3400 


2300 


2900 


White oak 


25 


7000 


3200 


5300 


Overcup oak 


45 


4900 


2800 


3800 


Cow oak 


58 


4900 


2300 


3800 


Texan oak 


39 


6000 


3100 


5200 


Willow oak . . 


49 


5500 


2300 


3800 


Spanish oak 


52 


5100 


2500 


3900 


Shagbark hickory .... 
Mockernut hickory . . . 
Water hickory 


22 
18 
4 


6900 
7200 
5600 


3500 
4500 
4700 


5700 
(5100 
5200 


Nutmeg hickory .... 
Pecan hickory 
Pignut hickory .... 


26 
4 
5 


5500 
3800 
6200 


3700 
3300 
4700 


4500 
3(500 
5400 


Sweet gum 


6 


3600 


3000 


3300 













Certain special tests were also made to determine : 

(a) The effect of bleeding (tapping for turpentine) on long-leaf pine. 

(b) Influence of size on transverse strength of beams. 

(c) Influence of size on compressive strength. 

(d) The effect of hot-air treatment in dry kilns on strength. 

The results obtained from these tests indicated : 
(a) That bleeding does not affect the strength. 



348 STRENGTH OF MATERIALS 

(&) That large, sound beams may be as strong as small ones cut 
from the same piece ; that is, large beams may show the same fiber 
stress as small ones. 

(c) That large, sound pieces in compression may be as strong as 
small ones cut from the same piece; that is, the intensity of com- 
pressive stress may be the same. 

(d) That there were no detrimental effects. 

The results of the tests made by the Bureau of Forestry, as out- 
lined in this article, should not be taken as conclusive, since not a 
sufficient number of tests were made to establish values. The pieces 
were in most cases small, and specially selected, and the results are 
of more value from a scientific than from a commercial standpoint, 
since the lumber of commerce contains knots, wind shakes, and other 
defects that lessen its strength. 

257. Recent work of the United States Forest Service. The United 
States Forest Service (formerly known as the Bureau of Forestry) 
has recently made extensive studies of the uses and durability of 
the various commercial woods of the United States, and has also 
conducted a series of tests to determine their strength, the most im- 
portant of which are as follows : 

() Tests of commercial-size beams of various timbers found on the market 
to determine 

1. The effect of knots and other defects on the strength. 

2. The effect of moisture on the strength. 

3. The effect of preservatives on the strength. 

4. The effect of methods of seasoning on the strength. 

(&) Tests of materials used in the construction for vehicles for such pur- 
poses as spokes, axles, and poles, 
(c) Tests of the strength of packing boxes. 
(d} Tests of the strength of railroad ties. 

In each of these investigations one of the objects has been to deter- 
mine, if possible, some so-called inferior woods that might be used in 
place of varieties that are superior but are becoming scarce. The test 
pieces for (a) were large commercial pieces in which knots and other 
defects occur, as they do in the structural timbers used by engineers. 

A summary of some of the cross-bending tests is given in the 
following table. 



TIMBEK 



349 



FLEXURE TESTS OF COMMERCIAL TIMBER 







, 


g 




WEIGHT PER 










11 


H 
. 


IB _^ 


CUBIC FOOT 


fa 
H 


8 E. 






H 


* 5 


J ^ 




J? ^ 


* 3 a 


SPECIES 


GRADE 


H C/2 


W fl 


H w 






2 H *~ 


iJ B \ 






**< ^ 


2 2 


22 ^ 


As 




D CM "^ 


Z> 






51 


a 

H 


I - 


Tested 
Ib. 


Dry 

Ib. 


a D 5 

tf 




Red flr : 


















ShipmentsA 


{Selects . . . 
Merchantable 


29 


Utol2 


r22.6 

^ 20.8 


37.1 
34.5 


30.2 

28.4 


8810 
7730 


1,925,000 
1,825,000 


and C 


Seconds . . 


16 


J 


U9.5 


31.9 


26.7 


6290 


1,630,000 




rSelects . . . 


14 


1 


p7.6 


30.9 


24.2 


6250 


1,280,000 


Shipment B . . 


-1 Merchantable 


15 


r 3 


^ 26.5 


33.7 


26.6 


5340 


1,320,000 




1 Seconds . . 


25 


J 


[26.2 


35.1 


27.8 


4280 


1,400,000 


ShipmentsA, 
B, and C . . 


rSelects . . . 
i Merchantable 
[.Seconds . . 


36 
44 
41 


} ' 


r24.5 
^22.7 
123.6 


34.7 
34.8 
33.8 


27.9 
28.4 
27.4 


7780 
6920 
5070 


1,675,000 
1,650,000 
1,490,000 


Average of 


















shipments A, 


















B, and C . . 


All grades 


121 


. . . 


23.6 


33.4 


27.8 


6580 


1,570,000 


Western hemlock 


All grades 


30 


3 to 6 


32.2 


35.4 


26.8 


5565 


1,260,000 


North Carolina 


















loblolly pine . 


Square edge . 


20 


3 


37.2 


42.8 


31.2 


6187 


1,479,000 


Long-leaf pine . 


Merchantable 


26 


6 to 12 


26.7 


53.3 


42.1 


8210 


1,790,000 



The grades selects, merchantable, and seconds, referred to in the table, 
are those from the Pacific Coast standard grading rules for Douglas 
Fir for 1900. A copy of these rules is here given. 

Merchantable. This grade shall consist of sound, strong lumber, free from 
shakes, large, loose, or rotten knots, and defects that materially impair its 
strength ; it shall be well manufactured and suitable for good substantial 
constructional purposes. 

Will allow occasional variations in sawing or occasional scant thicknesses, 
sound knots, pitch seams, and sap on corners, one third the width and one half 
the thickness. Defects in all cases to be considered in connection with the 
size of the piece and its general quality. 

Seconds. This grade shall consist of lumber having defects which exclude 
it from grading as merchantable. 

Will allow knots and defects which render it unfit for good substantial con- 
structional purposes, but suitable for an inferior class of work. 

Selects. Shall be sound, strong lumber, good grain, well sawn. 

Will allow, in sizes 6 by 6 and less, knots not to exceed 1 in. in diameter; 
sap on corners one fourth the width and one half the thickness ; small pitch . 
seams when not exceeding 6 in. in length. 



350 



STRENGTH OF MATERIALS 



114 



CCrf 



00 >* 



O a, 
S C 



m 

cr. o 



x~ 



51^ 



f ^V- ~ 

; ^ x i 



1.2 



JC^ 



00 5<1 . CD 00 C<1 . OOOCO . 00 C<1 . O 00 O5 . OOC<I . OOCO . r< T)H 00 00 C<1 . Clib 
XX .XXX .XXX .XX .XXX .XX .XX .XXXXX .XX 



Sr-' 



feo S 






858 



11 



0005 
X X 



X X 

5oci 

X X . 



3 o 

* .2 
-a -S 



.2 



^H r^ 

a 5 -a 

S S a 



S'o 



.S O .X 



" 25 * il 



O o 

I J 



as a 
a 

'a H 

t>, O 



s o -3 o ^3 o 
I r2 -- n "T< 






11 



3*1 



TIMBER 351 

In sizes over 6 by 6, knots not to exceed 2 in. in diameter, varying according 
to the size of the piece ; sap on corners not to exceed 3 in. on both face and edge ; 
pitch seams not to exceed 8 in. in length. 

Defects in all cases to be considered in connection with the size of the piece and 
its general quality. 

The cross-bending tests of 1 were made upon large specimens 
ranging in size from 6 in. x 8 in. x 7 ft., to 8 in. x 16 in. x 16 ft. The 
table shows that the modulus of rupture is less for the poorer grades 
of timber than for the selects, showing the effect of knots and other 
imperfections. The modulus of elasticity, indicating the stiffness, is 
less for the poorer grades, except in the case of shipment B of red fir. 

The same report also makes a comparison of the strength of large 
sticks and small sticks, both in cross bending and in compression 
parallel to the fiber. 

The table on page 350 gives average values obtained from this 
report, and indicates that the strength of the small sticks is, in nearly 
every case, greater than the strength of the large sticks. The modu- 
lus of elasticity is less for the small sticks than for the large ones, 
indicating a greater stiffness for the latter. 

258. Treated timber. The increasing scarcity of good timber and 
the consequent rise in price has called the attention of American 
engineers to the necessity for the use of preservatives in order to 
lengthen the life of the timber for commercial purposes. This has 
developed a new branch of engineering in this country, based on the 
use of many things learned by the Europeans, who were the originators 
of some of the best methods of treatment. 

When the tree is cut down and the timber seasoned (dried), a por- 
tion of the water evaporates from the sap, leaving the food materials 
deposited upon the cell walls. These materials are excellent food for 
bacteria and various forms of fungi that cause early decay of the 
timber if allowed to carry on their destructive work. In the early 
days, when timber was plentiful, no attempt was made to preserve 
wood from the destructive action of bacteria, but with increasing 
scarcity of good timber various methods of treatment have been 
devised. The simplest method, of course, is the application of com- 
mon paint. This closes all the pores and protects the wood from the 
action of bacteria, but this method cannot be made use of where the 



352 STRENGTH OF MATERIALS 

timber is in or near the ground or water, since the continued mois- 
ture causes the paint to peel off. The methods most generally used 
for treating timber for commercial purposes are given in the follow- 
ing paragraphs. 

Zinc chloride process. The zinc chloride process is the cheapest, and 
until the last few years the one most widely used in this country. 
It consists of impregnating the wood fibers with a solution contain- 
ing about one half a pound of dried zinc chloride per cubic foot of 
timber. The treatment is carried out as follows : Air-seasoned timber 
or timber that has been steamed to drive off the moisture is placed 
in a cylinder ; a vacuum is then maintained, while the solution is be- 
ing introduced, until the timber is covered. Pressure is then applied 
up to 100 to 125 lb./in. 2 by pumping in additional solution. When 
the penetration has been sufficient, the solution is drained off. The 
principal difficulty with the timber treated by this process comes from 
the injury caused by steaming and the subsequent rapid leaching out 
of the zinc chloride. This treatment requires about seven hours. 

Absorption process. In this process of treatment and those that fol- 
low the preservative used is creosote oil. This oil is obtained from 
coal tar, a by-product of artificial gas manufacture and the coke ovens. 
The creosote oil is distilled from coal tar at temperatures between 
240 and 270 C. This absorption process is also known as a non- 
pressure process. Air-dried timber is placed in a receptacle and cov- 
ered with the boiling preservative. This boiling tends to expel some 
moisture from the wood. After boiling, the excess creosote is drained 
off and the timber is immersed in cold preservative. In this way 
greater absorption is obtained on account of differences in tempera- 
ture and pressure. This process is used principally for butts of tele- 
graph poles, fence posts, and ties, in limited numbers. About 6 to 12 Ib. 
of creosote oil per cubic foot may be absorbed by this process. The 
time required for treatment varies from seven to fourteen hours. 

Full-cell creosoting process. The seasoned timber, which may be 
steamed to reduce moisture and expel sap, is placed in a vacuum and 
creosote introduced until the timber is submerged. A pressure of 
100 to 125 lb./in. 2 is then maintained, forcing the creosote into the 
wood. The creosote is then drained from the tank and, generally, 
a low vacuum is maintained to draw out the excess preservative. 



TIMBER 



353 



An absorption of as much as 20 Ib. of creosote oil per cubic foot 
of timber is possible by this process. The time required, including 
steaming, is about seven hours. 

Riiping process. When this treatment is used, compressed air is 
forced into the pores of the wood, and while under this compres- 
sion, creosote oil is introduced under a higher pressure (150 lb./in. 2 ). 
When the pressure is relieved and the creosote drained off, a vacuum 
is produced, allowing the compressed air in the pores of the wood to 
expand and force out the excess creosote. This leaves about 4 to % 6 Ib. 
of creosote per cu. ft. of timber. The cell walls are left lined with the 
preservative, whereas in the full-cell process the cells themselves are 
left nearly full. The Iviiping process is accordingly much more eco- 
nomical in the use of cresote. The time required for this treatment 
is about four hours. 

259. Strength of treated timber. The question naturally arises as 
to whether or not the treatment to which timber is subjected in in- 
troducing the preservative has any effect upon its strength in tension, 
bending, compression, and shear. The question as to whether or not 
the preservative itself weakens the timber must also be considered. 
To answer these questions the United States Forest Service has made 
an extended study of the strength of treated timber. The results of 
some of these tests are shown in the following tables. 



SOUTHERN-PlNE BRIDGE STRINGERS, TREATED AND UNTREATED 
STATIC BENDING, $ POINT LOADING 
NOMINAL SIZE, 8 IN. x IG IN. x 14 IN. 







Natural 


Treated 


Natural 


Treated 


Species 


Moisture 
Condition 


Deflection 
at Maxi- 


Modulus 
of 


Deflection 
at Maxi- 


Modulus 
of 


Fiber 
Stress at 


Fiber 
Stress at 






mum Load 


Rupture 


mum Load 


Rupture 


Elastic 
Limit 


Elastic 
Limit 






in. 


lb./in.2 


in. 


Ib./in.z 


lb./in.2 


lb./in .2 


Pines : Long-leaf 


air dry 


2.14 


6466 


1.64 


6376 


741 


793 


Loblolly 


air dry 


1.76 


6392 


1.47 


6380 


653 


461 


Long-leaf 


partially 


1.80 


5151 


1.90 


5132 


719 


586 




air dry 














Loblolly 


partially 


1.46 


4858 


1.60 


4150 


574 


352 




air dry 















COMPRESSION PER- 
PENDICULAR TO 

GRAIN 
NOMINAL SIZE 

8 IN. x 16 IN. x,30 IN. 



354 



STRENGTH OF MATERIALS 



BRIEF SUMMARY OF RESULTS OF TESTS ON TREATED TIES 



STRENGTH OF FULL-SIZED 
TIES ix RAIL BEARING 
(COMPRESSION PERPENDIC- 
ULAR TO GRAIN) 


COMPRESSIVE 
STRENGTH AT 
ELASTIC LIMIT 

lb./in.2 


PULLING RESISTANCE OF 
SPIKES 


LATERAL 
RESISTANCE 
OF SPIKES 

LOAD AT j-IN. 

DISPLACEMENT 
lb. 


Common 
lb. 


Screw 
lb. 


Natural red oak ties . . . . 


1093 


8026 


13,855 




Burnettized red oak ties . . 


1065 


7826 


13,781 




The above ties were from 










Carbondale, 111. 










Natural red oak ties .... 


1239 


8935 


14,686 




Creosoted red oak ties (Lowry) 
The above red oak ties 


1285 


8303 


14,522 




were from Shirley, Ind., 










and were drier than those 










from Carbondale, 111. 










Natural red oak ties . . . . 


1060 


6792 


11,418 


4028 


Treated red oak ties (Raping) 


978 


6299 


10,962 


4211 


Natural red oak ties .... 


962 


6805 


12,521 


4610 


Treated red oak ties (Full Cell) 


999 


6853 


11,671 


4458 


The above ties were from 










Somerville, Tex. 










Natural loblolly pine ties . . 


510 


3720 


8,003 




Treated loblollv pine ties 










(Ruping) 


503 


4583 


7,787 




Natural loblolly pine ties . . 


479 


3621 


8,200 




Treated loblolly pine ties 










(Lowry) 
The above loblolly pine 


499 


3407 


7,936 . 




ties were from Grenada, 










Miss. 










Natural loblolly pine ties . . 


619 


3660 


8,040 


2831 


Treated loblolly pine ties 










(Ruping) 


696 


3980 


9,020 


3486 


Natural loblolly pine ties . . 


730 


4755 


9,012 


2979 


Treated loblolly pine ties 










(Full Cell) 


729 


3986 


8,747 


3830 


Natural loblolly pine ties . . 


721 


3894 


8,911 


3200 


Treated loblolly pine ties 










(Crude Oil) 


529 


2069 


7,495 


2875 


The above loblolly pine 










ties were from Somerville, 










Tex. 










Natural short-leaf pine ties. . 


799 


4624 


11,136 


3525 


Treated short-leaf pine ties 










(Ruping) 


823 


4626 


9,684 


3645 


Natural short-leaf pine ties. . 


609 


4387 


9,714 


3254 


Treated short-leaf pine ties 










(Full Cell) 


659 


4532 


9,805 


3417 


Natural short-leaf pine ties . . 


517 


4068 


9,182 


3372 


Treated short-leaf pine ties 










(Crude Oil) 


373 


1816 


7,182 


3439 


Natural long-leaf pine ties . . 


677 


4465 


9,001 


3255 


Treated long-leaf pine ties 










(Ruping) 


737 


4458 


9,170 


3276 


Natural long-leaf pine ties . . 
Treated long-leaf pine ties 


703 


3445 


8,474 


3258 


(Full Cell) 


712 


3634 


9,290 


3542 


Natural red gum ties .... 


916 


4383 


10,010 


3789 


Treated red gum ties (Ruping) 


884 


4490 


9,720 


3861 


Natural red gum ties .... 


843 


3650 


9,565 


3765 


Treated red gum ties (Full Cell) 


833 


3815 


9,205 


3679 


Natural red gum ties .... 


731 


3615 


9,885 


3450 


Treated redgumties(CrudeOil) 


655 


2540 


9,750 


3500 


The above short-leaf and 










^ng-leaf pines and red 










gum ties were from Somer- 










ville, Tex. 











TIMBER 355 

An examination of the results of tests of the bridge stringers shows 
that there is little decrease in strength due to the action of the creo- 
sote in the case of the air-dry, long-leaf pine. The loblolly pine, air 
dried, shows a decrease in strength of 16 per cent in bending and 
29 per cent in compression. The long-leaf pine, partially air dried, 
shows no appreciable decrease in strength in bending, but about 18 
per cent decrease in compression. Loblolly pine, partially air dried, 
shows 14 per cent decrease in bending strength and 38 per cent in 
compressive strength. These tests seem to show that long-leaf pine 
is injured very little, if any, by the creosote, while loblolly pine is 
injured appreciably. Treated oak ties (results not given here) show 
a decrease in strength of from 5 to 10 per cent. Douglas fir and 
Wisconsin white pine show little or no effect due to treatment so far 
as bending and compression are concerned, but show a decrease in 
strength of from 20 to 25 per cent in shear. 



CHAPTER XVII 

ROPE, WIRE, AND BELTING 

260. Wire. Wire is made from a steel or iron rod by pulling it 
through a hole, or die, of smaller diameter than the rod. This is 
called drawing, and is done while the metal is cold. It is known as 
wet drawing when the metal is lubricated, and as dry drawing when 
no lubricant is used. The drawings are made with a smaller sized die 
each time, until the desired diameter of wire is obtained. Cold draw- 
ing of steel and iron raises the elastic limit and ultimate strength of 
the metal and decreases its ductility. It is made ductile again by 
annealing, and is finished by giving it the proper temper consistent 
with the desired use. 

The Mining Journal for 1896 gives the following values for the 
strength of wire. 

lb./in.* Ib./in- 2 

Iron wire 80,000 High-carbon steel wire . . 180,000 

Bessemer steel wire . . . 90,000 Crucible cast steel . . . 240,000 
Mild open-hearth steel wire 130,000 

Piano wire varies in strength from 3 00,0 00 lb./in. 2 to 400,0001b./in. 2 

261. Wire rope. Wire rope is made by twisting a number of steel 
or iron wires into a strand, and then twisting a number of these 

strands about one of the strands, or about a hemp, 
manila, jute, or cotton strand. The exact composition 
of the cable or wire rope will depend upon the service 
for which it is designed. The hemp core gives added 
pliability to the cable, and acts as a means of lubricat- 
ing the strands and wires ; this reduces the internal 
friction in the cable, and adds much to its life in case it is used where 
pliability is required, as in running over sheaves. Fig. 194 is an illus- 
tration of the cross section of a cable in which the separate strands 
each have a hemp core. Such a cable can be used where great 
pliability is required. Fig. 195 shows a cross section of a cable with 

356 






KOPE, WIKE, AND BELTING 357 

a single hemp core at the center, and Fig. 196 shows a cross section 
of a cable in which the center is a wire strand similar to those used 
on the outside. A cable of the latter type can only be used where 
little bending is required, as in 
the case of suspension bridges. 
The strands are twisted about the 
central core either to the right or 
left. When twisted to the left 

the rope is designated as left lay, 

FIG. 195 FIG. 196 

and when twisted to the right as 

right lay. The twist is long or short, depending upon the require- 
ments of service. The shorter the twist the more flexible the rope, 
and the longer the twist the less flexible. 

262. Testing of rope wire and belting. These materials are usually 
tested in tension. This may be done in an ordinary testing machine, 
providing the proper means are used for holding the specimen. A 
type of wire-testing machine is shown in Fig. 197. One end of the 
wire is clamped to the movable head and the other to the stationary 
head, which is provided with a spring balance for registering the pull. 
Many other types of wire-testing machines are in use, some of them 
being arranged to make torsion tests. Many special machines are 
also made for testing rope and belting. 

Since a wire rope is a built-up structure, made of twisted strands, 
it is not to be expected that it will exhibit such well-defined elastic 
properties as a single wire tested separately. This is due to the fact 
that as the tension is increased each strand, which was originally in 
the form of a helix of a certain pitch, becomes somewhat straightened 
and takes the form of a helix of a greater pitch. On account of the 
twisted condition of the wires in the strands, they do not all carry 
the same load, and therefore do not all reach their elastic limit at 
the same time. We find, consequently, upon testing a wire rope, that 
it has no well-defined elastic limit. 

The individual wires of which the rope is made show a very high 
tensile strength and elastic limit, but exhibit no yield point, as the 
process of drawing seems to destroy the properties of the material 
that give the yield-point phenomena. The modulus of elasticity is 
not changed appreciably by the process of drawing. 



358 



STEENGTH OF MATERIALS 



Problem 315. A piece of steel music wire was tested in tension and the following 
data obtained. Draw the strain diagram, using loads in lb./in. 2 as ordinates and 
unit elongations as abscissas, and find the elastic limit, the modulus of elasticity, 
and the modulus of elastic resilience. The wire was No. 25 gauge ; diameter before 
test 0.0577 in., and sectional area 0. 002615 sq. in. It was tested on a gauge length 
of 6 in. The sectional area at the point of fracture after test was 0.00132 sq. in. 
Compute the percentage of reduction of cross section. 

TEST OF WIRE 



LOAD 
Ib. 


ELONGATION 
in. 


LOAD 

Ib. 


ELONGATION 
in. 


100 


.0058 


660 


.0627 


200 


.0146 


680 


.0661 


300 


.0223 


700 


.0698 


400 


.0316 


720 


.0752 


500 


.0415 


740 


.0791 


520 


.0450 


760 


.0852 


540 


.0463 


780 


.10936 


560 


.0489 


800 


.1039 


580 
600 


.0512 
.0546 


f Tensile strength, 
\ 320,460 lb./in.2 


620 


.0564 




640 


.0591 





263. Strength of wire rope. The following report of tests of steel 
rope is taken from the Watertown Arsenal Report, 1889. 



TENSION TESTS OF STEEL WIRE ROPE 



CIRCUMFER- 


NUMBER 


WIRES 


MEAN 
DIAMETER 




SECTIONAL 


TENSILE STRENGTH 


ENCE 
in. 


OF 

STRANDS 


PER 

STRAND 


OF WIRES 


CORE 


WIRE 


Total 


Total 








in. 




in .2 


Ib. 


lb./iu. a 


1.5 


6 


18 


.0321 


Hemp 


.0876 


12,898 


147,236 


1.75 


6 


18 


.0349 


it, 


.1031 


15,736 


153,893 


2 


6 


18 


.0420 


it 


.1499 


20,780 


138,360 


2.125 


6 


18 


.0456 





.1766 


24,430 


138,383 


2.25 


6 


18 


.0488 


u 


.2021 


30,960 


148,650 


2.50 


6 


18 


.0544 


u 


.2510 


33,270 


132,500 


3 


6 


18 


.0598 


u 


.3024 


46,370 


153,340 


3.50 


6 


18 


.0718 


(( 


.4380 


65,120 


148,675 


4.50 


6 


18 


.0980 


u 


.8151 


138,625 


170,075 



ROPE, WIRE, AND BELTING 



359 



TEST OF INDIVIDUAL WIRES TAKEN FROM THE WIRE ROPE 
REPORTED ABOVE 





DIAMETER OF 


SECTIONAL 


TENSILE STRENGTH 


SIZE OF ROPE 


WIRE 


AREA 




in. 


in. 


in.2 


Ib. 


Ib./in.z 


1.50 


.0325 


.00082 


130 


158,540 


2.00 


.0430 


.00145 


226 


155,860 


2.50 


.0546 


.00234 


502 


214,530 


2.75 


.0593 


.00276 


452 


163,770 


3.00 


.0600 


.00283 


478 


168,900 


3.50 


.0725 


.00413 


594 


143,830 


4.50 


.9980 


.00754 


1390 


184,350 



In the above table the actual sectional area of the wire in the rope 
is given, and the tensile strength in lb./in. 2 has been computed by 
dividing the total load by this area. An examination of the table 
giving the strength of the individual wires shows that the intensity 
of stress is greater in the case of the individual wires than in the wire 
rope ; that is to say, the structure of the rope causes the wires to lose 
some of their efficiency. 

STRENGTH OF IRON WIRE ROPE AS GIVEN BY JOHN A. ROEBLING 
(Rope composed of six strands and a hemp center, seven or twelve wires in each strand) 



DIAMETER 


CIRCUMFERENCE 


APPROXIMATE BREAK- 
ING STRENGTH 


CIRCUMFERENCE 
IN INCHES OF NEW 


in. 


in. 




MANILA ROPE 






Ib. 


OF EQUAL STRENGTH 


1.75 


5.50 


88,000 


11 


1.625 


5.00 


72,000 


10 


1.50 


4.75 


64,000 


9.6 


1.375 


4.25 


52,000 


8.5 


1.25 


4.00 


46,000 


8.0 


1.125 


3.50 


36,000 


6.5 


1.000 


3.00 


26,000 


6.76 


.875 


2.75 


22,000 


6.25 


.750 


2.25 


14,600 


4.75 


.500 


1.50 


6,400 


3.00 


.376 


1.125 


3,600 


2.25 


.250 


.75 


1,620 


1.50 



360 



STRENGTH OF MATERIALS 



The table at the bottom of page 359 gives the strength of iron and 
cast-steel wire rope as given by John A. Roebling's Sons. The size of 
a new manila rope of the same strength is also given for comparison. 



STRENGTH OF WIRE ROPE MADE FROM CAST STEEL AS GIVEN 
BY JOHN A. ROEBLING 

(Rope composed of six strands and a hemp center, seven or nineteen wires in each strand) 



DIAMETER 


CIRCUMFERENCE 


APPROXIMATE BREAK- 
ING STRENGTH 


CIRCUMFERENCE 
IN INCHES OF NEW 


in. 


in. 




MANILA ROPE 






Ib. 


OF EQUAL STRENGTH 


1.25 


4.00 


106,000 


13 


2.125' 


3.50 


82,000 


11 


1.00 


3.00 


62,000 


9 


.875 


2.75 


52,000 


8.5 


.750 


2.25 


35,200 


7.0 


.025 


2.00 


28,000 


6.0 


.500 


2.50 


16,200 


4.75 


.375 


1.125 


9,000 


3.75 



Problem 316. A wire cable of the following dimensions and composition was 
tested, and its maximum load found to be 5080 Ib. Diameter of cable, 0.33 in. ; six 
strands of eleven wires each; sectional area of wires, 0.0253 in. 2 A test of the 
individual wires showed an average strength of 225,600 lb./in. 2 Find the loss of 
strength due to the twisting of the wires to form the cable, assuming that all the 
wires have the average strength given above. 

264. Strength of manila rope. The following table gives the 
strength of manila and sisal rope as computed from tests made by 
the Watertown Arsenal.* The load in lb./in. 2 is given in each case. 
This has been computed by considering the cross section of the rope 
as the area of a circle of the same diameter. It will be seen from 
the table that the stress for the smaller ropes was 15,000 lb./in. 2 , 
while for the larger ropes it was only about 7000 lb./in. 2 This dif- 
ference is due in part to the greater length of yarn used in the 
smaller rope. Manila rope has about two thirds the strength of 
good Eussian hemp rope.f The United States Navy test allows 
1700 lb./in. 2 as the working strength of a 1.75-in. hemp rope. 



Watertown Arsenal Report, 1897. 



t Thurston, Materials of Construction. 



ROPE, WIRE, AND BELTING 



361 



TESTS OF MANILA AND SISAL ROPE 
MANILA ROPE 



SIZE OF ROPE 


DIAMETER 
in. 


SECTIONAL, 
AREA 

in.2 


TENSILE STRENGTH 


TOTAL 
LOAD 

Ib. 


lb./in.2 


Per Yarn 
Ib. 


6-th read 
9-th read 


.27 
.30 
.38 
.43 
.49 
.56 
.61 
.62 
.74 
.79 
.78 
.85 
.96 
1.00 
.99 
1.13 
1.19 
1.29 
1.28 
1.39 
1.34 
1.41 
1.59 
1.61 
1.66 
1.76 
2.25 
2.52 
2.83 
3.35 
3.70 


.0567 
.0750 
.114 
.153 
.192 
.259 
.288 
.299 
.41 
.478 
.462 
.557 
.715 
.782 
.746 
.970 
1.07 
1.27 
1.26 
1.46 
1.36 
1.51 
1.88 
1.99 
2.04 
2.35 
3.82 
4.86 
6.22 
8.37 
10.06 


13,360 
14,180 
12,920 
14,250 
11,610 
11,970 
10,800 
11,500 
9,200 
12,900 
11,900 
12,470 
12,810 
13,630 
13,750 
12,470 
12,190 
11,990 
11,610 
10,080 
11,790 
9,890 
10,360 
10,480 
10,740 
9,940 
8,260 
9,400 
8,600 
7,500 
7,300 


126 
118 
123 
145 
125 
148 
130 
128 
114 
148 
138 
136 
153 
146 
151 
144 
132 
134 
132 
117 
130 
113 
121 
134 
128 
118 
112 
125 
118 
108 
102 


750 

1,064 
1,473 
2,180 
2,242 
3,100 
3,120 
3,455 
3,775 
6,207 
5,509 
6,947 
9,160 
10,663 
10,260 
12,093 
13,050 
15,227 
14,640 
14,723 
16,017 
14,943 
19,577 
20,873 
21,903 
23,360 
31,570 
45,647 
54,000 
62,717 
73,910 


12-thread 
15-thread . . . 


1.25-in 


1.50-in 
1 625-in 


1 75-in 


2-in 


2 25-in 


2 25-in. 


2 50-in 


2.75-in 


3-in. . . 




3 25-in 


3 50-in 


3 75-in. . 


3 75-in. 


4-iu. 


4-in 


4.25-in 
4 50-in 


4 50-in. . ... 


4.75-in 
5-in 


6-in 


7-in 


8-in. 


9-in 


10-in . . 





SISAL ROPE 



SIZE OF ROPE 


DIAMETER 
in. 


SECTIONAL 
AREA 

in.2 


TENSILE STRENGTH 


TOTAL 
LOAD 

Ib. 


lb./in.2 


Per Yarn 
Ib. 


6-thread 


.27 
.33 
.39 
.45 
.56 
.63 
.70 
.81 
.95 
1.01 
1.22 


.0567 
.082 
.126 
.129 
.254 
.302 
.395 
.416 
.691 
.780 
1.128 


7,700 
7,300 
7,500 
10,810 
8,100 
7,600 
7,200 
9,500 
8,300 
7,500 
7,200 


72 
67 
79 
93 
99 
96 
97 
94 
101 
104 
102 


432 

605 
944 
1397 
2067 
2315 
2925 
3966 
5733 
5917 
8230 


9-thread ... . . 


12-thread . . 


1 25-in 


1 50-in ... 


1 75-in. . 


2-in 


2.25-in 
2.75-in 


3-in. ... 


3.50-in 





362 



STRENGTH OF MATERIALS 



265. Strength of leather and rubber belting. Leather belts are 
made from tanned oxhide. That portion of the hide that originally 
covered the back gives the best leather for this purpose. The " flesh 
side," or side originally next to the animal, wears better when placed 
in contact with the pulley, while the outside gives the greater 
adhesion when placed in contact with the pulley. 

Single belts are made from one thickness of leather, the desired 
length being obtained by cementing or splicing the short lengths cut 
from the hide. Double belts are made by cementing two thicknesses 
of the leather together. The strength of good leather varies from 
600 to 700 Ib. per inch of width, and from one half to two thirds 
as much when spliced. The following table gives the strength of 
cemented belt laps as determined by the Watertown Arsenal.* A 
complete series of tests on belt lacings is also reported in the same 
volume, and the student is referred to this report for the results. The 
allowable stress on a single belt is from 250 to 300 Ib. per inch 
of width. 



TESTS OF LEATHER BELTING 





DIMENSIONS 












TENSILE STRENGTH 




in. 


SECTIONAL 




DESCRIPTION t 




AREA 










Thick- 


in.* 




Pounds per 




Length 


Width 






lb./in.2 


Inch of 








ness 






Width 


2-in., single . . . 


60.00 


1.98 


.20 


.396 


5045 


1091 


6-in., single . . . 


60.20 


6.07 


.22 


1.34 


2537 


560 


6-in., single (w) 


60.11 


6.08 


.24 


1.46 


2119 


533 


12-in., single . . . 


60.11 


12.05 


.18 


2.17 


3917 


705 


4-in., double . . 


59.55 


3.98 


.33 


1.31 


4931 


1623 


6-in., double . . 


60.18 


6.91 


.47 


2.78 


4309 


2027 


6-in., double (w) . 


59.93 


6.00 


.40 


2.40 


5166 


2066 


12-in., double . . 


59.90 


11.90 


.39 


4.64 


4090 


1595 


12-in., double (w) . 


60.06 


11.93 


.36 


4.29 


4424 


1591 


24-in., double (w) . 


60.00 


23.90 


.47 


11.23 


2760 


1297 


30-in., double 


59.90 


29.95 


.43 


12.88 


2717 


1169 



* Watertown Arsenal Report, 1803. 

t The letter w iu the table stands for waterproofed. 



HOPE, WIKE, AND BELTING 



363 



TESTS OF RUBBER BELTING 





DIMENSIONS 








in. 


SECTIONAL 


TENSILE STRENGTH 


DESCRIPTION 




AREA 










Thick- 


in. 2 


j Pound per 




Length 


Width 


ness 




lb./in. 


Inch of 














Width 


2-in., 4-ply . 


60.17 


2.02 


.26 


.525 


3276 


851 


6-in., 4-ply . . . 


60.17 


6.08 


.26 


1.58 


3227 


839 


6-in., 4-ply . . . 


60.12 


6.13 


.26 


1.59 


3773 


979 


6-in., 4-ply . . . 


60.17 


6.05 


.26 


1.57 


2739 


711 


12-in., 4-ply . . . 


60.02 


12.08 


.27 


3.26 


3037 


819 


12-in., 4-ply . . . 


60.14 


12.24 


.26 


3.18 


2987 


776 


2-in., 6-ply . . . 


60.17 


2.14 


.36 


.770 


3101 


1116 


6-in., 6-ply . . . 


59.98 


6.26 


.37 


2.32 


2737 


1014 


6-in., 6-ply . . . 


60.08 


6.27 


.36 


2.26 


3770 


1358 


12-in., 6-ply . . . 


60. 15 


12.04 


.36 


4.33 


3436 


1236 


12-in., 6-ply . . . 


60.17 


12.16 


.34 


4.13 


3862 


1311 


24-in., 6-ply . . . 


60.13 


24.11 


.41 


9.89 


2381 


977 


30-in., 6-ply . . . 


60.04 


30.18 


.40 


12.07 


2808 


1123 



ANSWERS TO PROBLEMS 



1. 


17. 7 lb./in. 2 


9. 


31,024 Ib. 


21. 


18*. 




2. 


3.1 lb./in. 2 


10. 


.000307 in. 


22. 


1| in. iron wire rope. 


3. 


s = .0018. 


11. 


.00095 in. 


40. 


320 lb./in. 2 , 




4. 


5.4 in. 


12. 


1005 Ib. 




19 19.8', 




5. 


.0000104. 


13. 


.24 in. square. 




109 19.8'. 




6. 


8 = .002122, 


14. 


i in. 


41. 


p' = 2000 lb./in. 2 , 






12. 73 in. 


15. 


99 tons. 




q' = 3460 lb./in. 2 , 




7. 


16,500, 000 lb./in. 2 , 


16. 


20.0086 ft. 




g'max = 4000 11). /ll! 


. 2 




approximately. 


17. 


1890 lb./in. 2 


42. 


10,000 lb./in. 2 




8. 


.0055 in., 


18. 


.16 in. 


43. 


58,435 Ib. 






approximately. 


19. 


13,320 lb./in. 2 


44. 


23,868 Ib. direct stress, 




8 = .00002546. 


20. 


1.7, approximately. 




77,748 Ib. shear. 




45. 


5656 lb./in. 2 




47. m = 


5. 






46. 


With a factor of safety 


of 


5, 48. p e = 


556 lb./in. 2 f or m = 3J-. 




d = 1.21 in. 




49. Pc = 


12,646 lb./in. 2 for m = 3. 


57 


bh s 


62. 


M!-. ^ 2 = 261.3: 149.3. 67. 


&(bh 3 -b'h' s ). 




t/ 1 


36' 


Cf> 


bh 2 bh 2 -rrd 3 


fin 


n-d 4 d 






TTCZ 4 


DO. 


6 24 32 


OU. 


p ~~ 32 ' p ~ o V 


Q 


58. 


~64' 


64. 


S l :S z = 5:2. 


76. 


Zero at center, 




60. 
61. 


35,350 ft. Ib. 
942,500 ft. Ib. 


66. 


1<3^). 




1500 Ib. at ends. 





77. 4250 Ib. and 4750 Ib. at ends, 2750 Ib. and 1750 Ib. between loads. 



78. 



At center ; 
1837.5ft. Ib. 



79. 408 lb./in. 2 

80. 14,603 lb./in. 2 



120. y = 



122. y = 



123. 



84. S = 66.46 in. 3 

85. In the ratio 1 
PP 

3 El' 



24 El 

w 



384^7 



124. 
125. 



D= .7 in. 
D = .67 in. 



127. D = 



Pds (l-d} 3 



SEI 
128. L 



3 El P 

129. D = .061 in. for E c = 2,000,000 lb./in. 2 
134. M v = J/g = 0, . J/ 2 = 3/ 5 = - T 2 ? ioi 2 , 



136. 



138. h = 7.86 in. 

170. 350 tons. 

171. 9^ in. square. 

172. 5.82 in. 



192 AV 



= 39.7in. Ib. 

173. 6| in. wide for angles | in. thick. 

174. Rankine 616 tons, Johnson 627 tons. 

175. Rankine 268 tons, Johnson 267 tons. 

176. Assume various lengths for the column. 

365 



366 



STRENGTH OF MATERIALS 



177. 127 tons. 

178. 15 + . 

179. 2f in. square. 

190. = 11,490,000 lb./in. 5 

191. M = 43.24 in. Ib. 

192. d = 4.465 in. 



193. d = 3.684 in. 

194. Internal diameter = 

5.63 in. ; 
solid : hollow = 3:1. 

195. 4484. 

196. p e = 23,500 Ib. /in. 2 



197. If weight of shaft is 

neglected, 
q = 131) Ib./ in. 2 , 
II = 2f 

198. d = 7.114 in. 

199. = 32 28'. 



201. Angle of twist per unit of length is 6 l = V 33.8". 



203. g max = 22,2401b./in. 2 , 
D = 6.36 in., 
W =158.965 in. Ib. 
2940 Ib./ in.' 2 
375 lb./in. 2 , assuming 

10 for the factor of 

safety. 



221. 



230. 591 lib./ in. 2 

231. 15,880ft. 

232. 79.4. 

233. 12,187 lb./in. 2 

249. 1.2 in. 

250. 2344 lb./in. 2 

252. 139 lb./in. 2 

253. .28 in. 



223. Bottom .13 in. ; 
side .31 in. 

224. 65281b./in.2 

225. fin. 

226. 685 lb./in. 2 

227. 68|. 

228. lin. 

229. .13 in. 

254. 11. 78 lb./in. 2 

255. Assuming E 8 : E c = 15 : 1, /' = 2350 in. 4 , 
t' = 2.266 in., p = 450 lb./in. 2 

270. p max = 3733 lb./in. 2 , factor of safety 13, d = .0245 in. 
272. Pmn = 192.6 lb./in. 2 
273. d= .0002in., 291. 3 in. 

3f = 1.529 in. Ib. 293. Weyrauch,40421b./ft. 

289. E = 300 tons, by (104); Rankine, 4116 Ib./ft. 

# = 32 7 tons, by (105). 294. 4242 Ib./ ft. 
303. 450 lb./in. 2 

310. (a) 12,870 lb./in. 2 , 13,059 lb./in. 2 ; 
(6) 2659 lb./in. 2 , 8372 lb./in. 2 ; 
(c) 1908 lb./in. 2 , 5538 lb./in. 2 
312. 752 lb./in. 2 



295. 13,890 Ib./ ft. 
300. 2250 lb./in. 2 , 
3091 lb./in. 2 
302. 476 lb./in. 2 



INDEX 



(The numbers refer to pages.) 



Abrasion test of stone, 329 
Absorption test of brick, 335 

of stone, 329 
Allowance for shrinkage and forced fits, 

169 
Angle of repose, 245 

of shear, 138 

of twist, 138, 139, 145 
Annealing, 12 
Annual rings, 336 
Answers to problems, 365, 366 
Antipole and antipolar, 67 
Arch, linear, 218 

Arched rib, continuous, fixed at both 
ends, 238 

graphical determination of linear 
arch, 234, 239 

method of calculating pole distance 
of, 233 

stress in, 230 

temperature stresses in, 236, 242 

three-hinged, 231 

two-hinged, 231 
Arches. See Masonry arches 

equilibrium polygon for, 210, 212, 

213, 214, 215, 216 
Area, contraction of, 14, 269 
Ash, strength of, 344, 346 
Average constants, Table I 

Bald cypress, strength of, 346, 347 

Basswood, strength of, 344 

Beams, bending moments, 38, 50, 51 

built-in, 86 

cantilevers, 57 

cast-iron, 280 

Castigliano's theorem, 103, 104 

continuous, 88, 104 

deflection of, 83, 85, 88, 111, 114 

designing of, 56 

eccentric loading of, 65 

effect of shear on elastic curve of, 86 

elastic curve of, 36, 81, 84, 87, 89, 91 

impact and resilience, 94 

influence line for bending moment, 
96 

influence line for reactions, 101 

influence line for shear, 98 

limitation to Bernoulli's assump- 
tion, 85 



Beams, maximum moments, 50, 52, 53, 

54,55 

Maxwell's theorem, 99 
modulus of rupture, 282, 334 
moment of resistance, 39 
moments of inertia, 38, 43, 45 
oblique loading, 64 
of considerable depth, 133 
principle of least work, 106 
reactions of supports, 49, 52, 54, 55, 

101 

straight-line law, 37, 320 
theorem of three moments, 90 
vertical shear, 49, 58, 60, 62 
work of deformation, 93 

Bearing power of soils, 243 

Beech, strength of, 344 

Behavior of iron and steel in tension, 
270 

Belting, strength of, 362, 363 

Bending, cold, 279 

Bending and torsion combined, 33 

Bending moment, defined, 38 
maximum, 50, 52, 53, 54, 55 

Bending moment and shear, relation 
between, 55 

Bernoulli's assumption, 36 

Bessemer process, steel manufacture, 288 

Black walnut, strength of, 344 

Bond between concrete and steel, 321 

Box elder, strength of, 344 

Brick, absorption of, 335 
compression of, 330, 331 
flexure of, 334 
manufacture of, 330 
modulus of elasticity of, 332 
rattler test of, 334 

Brick piers, strength of, 331 

Briquettes, cement, 300 

compression of halves of, 302 
molding and care of, 301 
tensile strength of, 302 

Building blocks, concrete, 311 

Bureau of Forestry timber tests, 345, 
346, 348 

Bursting pressure of thick cylinder, 165 

Cantilever, 57 
Carbon, in cast iron, 280 
in steel, 289 



367 



368 



STRENGTH OF MATERIALS 



Cast iron, manufacture and general 
properties of, 279 

compression of, 282 

elasticity of, 282 

flexure of, 282 

impurities in, 280 

malleable, 285 

modulus of rupture of, 286 

shear of, 282 

specifications for, 285 

tensile strength of, 280, 286 
Cast-iron columns, 284 
Castigliano's theorem, 103 

application to continuous beams, 104 
Castings, malleable, 285 

steel, 291, 292, 293 
Cedar, strength of, 344, 346, 347 
Cement, 297 

compression tests of, 302, 303 

specifications for, 303 

test of fineness, 299 

test of soundness, 298 

test of tensile strength, 300, 302 

test of time of setting, 299, 300 
Cinder concrete, 310 
Circular plates, 179, 181 
Circular shafts in torsion, 138, 139, 140 
Classification of materials, 9 
Coefficient of cubical expansion, 29 
Coefficient of elasticity. See Modulus of 

elasticity 

Coefficient of linear expansion, 12 
Cold bending test, 279 
Column footings, 251 
Columns : 

cast-iron, 284 

Cooper's modification of Johnson's 
straight-line formula, 132 

eccentrically loaded, 133 

Euler's formula, 122, 123 

Gordon's formula, 126 

independent proof for fixed ends, 
123 

Johnson's parabolic formula, 128, 
129 

Johnson's straight-line formula, 
130, 131, 132 

modification of Euler's formula, 125 

nature of compressive stress, 120 

one or both ends fixed, 122 

Rankine's formula, 126, 127 
Combined bending and torsion, 33, 141 
Common theory of flexure,~3~6 
Compression, defined, 2 

brick in, 330, 331, 332 

brick piers in, 331 

cast iron in, 284 

cement in, 302, 303 

concrete in, 308, 309, 310 

stone in, 327, 328 



Compression, tests, 269 

timber in, 338, 344, 346, 347, 350 

Compressive strength, average values, 
Table I 

Concrete. See Masonry arches 
building blocks of, 311, 312 
mixing, 307 

modulus of elasticity of, 309 
reenforced. See Reenforced concrete 
tests of, 305, 306, 307, 308, 309, 310 

Concrete-steel plates, 187 

Consequence of Bernoulli's assumption, 
37 

Continuous beams. See Beams 

Contraction of area, 14, 271 

Core section, 67, 68, 69 

Cottonwood, strength of, 344 

Crane hook, design of, 205 

Cross-bending. See Flexure 

Crushing. See Compression 

Curvature due to bending moment, 36 

Curve, elastic. See Elastic curve 

Curved pieces, 191 

Cylinders and spheres, thin, 154 

Cylinders, thin : 

elastic curve for, 157 
hoop tension in, 155 
longitudinal stress in, 155 

Cylinders, thick, 162 

bursting pressure, 165 
Lamp's formulas, 162 
made of concentric tubes, 166 
maximum stress in, 164, 165 

Cypress, strength of, 346, 347 

Dangerous section, 51 

Deflection, of beams. See Beams 
of columns. See Columns 
bending, general formula, 111 
shearing, general formula, 114 

Deformation, defined, 2, 4 

Designing of arches, 227 

Designing of beams, 56 

Diagram, bending moment and shear, 51 

Douglas spruce, strength of, 346 

Earth pressure (retaining walls), 253 

Eccentric loading, 65 

Efficiency of riveted joint, 172 

Elastic afterwork, defined, 10 

Elastic constants, relation between, 30 

Elastic curve, 36, 81, 84, 87, 89, 91 

Elastic law, 8 

Elastic limit, defined, 6, 274 

Elastic resilience, 94, 274 

Ellipse of inertia, 47 

Ellipse of stress, 26 

Elliptical plates, 182, 183 

Elliptical shafts, 144 

Elm, strength of, 344, 346 



INDEX 



369 



Empirical formulas (arches), 227 
Equilibrium polygon (arches), 210, 212, 

213, 214, 215, 216 
Equivalent stress, 31 
Euler's formulas, 122, 123, 125 
European tests of timber, 343 
Expansion, cubical coefficient of, 29 
linear coefficient of, 12 

Factor of safety, 16, Table I 

Fatigue of metals, 10 

Fir, red, strength of, 344, 349, 350 

Flat plates. See Plates 

Flexural deflection, general formula 
for, 111 

Flexural rigidity, 148 

Flexure, common theory of, 34 

tests in, 269, 282, 312, 329, 334, 340, 
344, 346, 349, 350 

Flow of material, 10 

Form of test piece, 13, 270, 294 

Foundation. See Retaining walls 

Fracture, character and appearance, 272 

Fraenkel formula for flexural deflec- 
tion, 112 

Fragility, 11 

Functions of angles, Table X 

Gordon's formula, 126 

Granite, strength of, 18 

Guest's formula for combined bending 

and torsion, 142 
Gum, strength of, 344, 346, 347 
Guns. See Thick hollow cylinders 
Gyration, radius of, 42 

Hardening effect of overstrain, 11, 271 

Heartwood and sapwood, 337 

Helical spring, 145 

Hemlock, strength of, 344, 349, 350 

Hemp rope, 360 

Hickory, strength of, 344, 346, 347 

High-speed steel, 288 

Holding tension specimens, 270 

Hollow cylinders. See Cylinders 

Hollow spheres. See Spheres 

Hooke's law, 6 

Hooks, links, and springs : 

analysis for hooks and links, 191 
bending strain in curved piece, 191 
curvature, sharp, effect on strength, 

200 
curved piece of rectangular cross 

section, 198 
maximum moment in circular piece, 

201 

plane spiral springs, 203 
simplification of formula, 194 

Hydraulic cement, 297, 298 

Hysteresis, 10 



Impact and resilience, 94 
Impact tests, 278 
Indentation test, 343 
Independent proof (columns), 123 
Inertia, ellipse of, 47 

moment of. See Moment of inertia 
Influence line, for bending moments, 96 

for reactions, 101 

for shear, 98 
Initial internal stress, 12 
Iron, cast. See Cast iron 

ingot, 287 

wrought. See Wrought iron 
Iron and steel, 265 

strength of, at high temperatures, 

272 
Ironwood, strength of, 344 

Johnson's parabolic formula, 128, 129 

Johnson's straight-line formula, 130, 131 

Cooper's modification of, 132 

Keep's tests of cast iron, 282 
Kirkaldy's tests, 265, 282 

Lamp's formulas, 162 

Latent molecular action, 11 

Lateral contraction, 14, 271 

Law of continuity, 21 

Least work, principle of, 106 

Leather belting, 362 

Lime, manufacture and properties, 297 

Limestone, 326, 327, 328, 329 

Limit of elasticity. See Elastic limit 

Limitation to Bernoulli's assumption, 85 

Linear arch, 218, 234, 239 

Linear strain, 25 

Linear variation of stress, 320 

Links, hooks, and springs. See Hooks 

Load line (arch), 217 

Logarithms, common, Table VIII 

conversion of, Table IX 

natural, Table IX 

Manganese in cast iron and steel, 280, 

289, 290 

Manila rope, 360 
Maple, strength of, 344 
Masonry arches, 216 

application of principle of least 
work, 224 

conditions for stability, 220 

designing of arches, 227 

empirical formulas, 227 

linear arch, 218 

load line, 217 

maximum compressive stress, 222 

Moseley's theorem, 222 

oblique proj ection of, 229 

stability of abutments, 229 



370 



STRENGTH OF MATERIALS 



Masonry arches, Winkler's criterion for 

stability, 225 

Materials .not obeying Hooke's law, 71 
Materials of construction, average con- 
stants for, Table I 

Maximum bending moment. See Bend- 
ing moment 

Maximum earth pressure, 253 
Maximum normal stress, 23 
Maximum shear, 25 

Maximum stress in circular shafts, 138 
Maxwell's theorem, 99 
Measure of strain, 31 
Merchantable timber, 349 
Modulus of elasticity : 

average values of, Table I 

denned, 6, 274 

of brick, 332 

of cast iron, 282 

of concrete, 310 

of shear, 29 

of steel, 293 

of stone, 332 

of timber, 344, 346, 349, 350 

of wrought iron, 293 
Modulus of resilience, 94, 274 
Modulus of rigidity, 29 
Modulus of rupture, 282, 329, 334, 338, 

344, 346, 349, 350 
Moisture in timber, 337 
Molds for cement briquettes, 301 
Moment diagrams, 51 
Moment of inertia, by graphical method, 
43 

defined, 38 

of non-homogeneous sections, 45 

polar, 41 

tables of, Tables VI and VII 
Moment of resistance, 39 
Mortar, cement. See Cement 

lime. See Lime 
Moseley's theory, 222 

Natural cement, 297 
Neutral axis, defined, 35 

of sections of beams with oblique 

forces acting, 64 
of sections of reenf orced concrete 

beams, 316, 319, 321 
Nickel steel, 290 

Non-circular shafts in torsion, 143 
Normal stress, maximum, 23 

Oak, strength of, 344, 346, 347 
Oblique loading, 64 
Oblique projection of arch, 229 
Open-hearth steel, 288 
Ordinary foundations, 249 
Overstrain, effect of, on iron and steel, 
11, 271 



Parabolic variation of stress, 318 

Paving brick. See Brick 

Phosphorus in iron and steel, 280, 290 

Physical constants, Table I 

Piers, brick, 331 

Piles, bearing power of, 246 

Pine, strength of, 344, 346, 347, 350 

Pitch of rivets, 172 

Planar strain, 22 

Plates, flat, circular, 179, 181 

concrete-steel, 187 

elliptical, 182, 183 

formulas of Bach, Grashof , Nichols, 
and Thurston, 190 

rectangular, 186 

square, 185 

stress in, 179, 181, 183, 185, 186 

theory of, 179 
Poisson's ratio, 7, Table I 
Polar moment of inertia, 41 
Poplar, strength of, 344 
Portland cement, 297 
Power transmitted by circular shafts, 

140 

Principal axes, 41 
Principal stresses, 24 
Principle of least work, 106, 224 
Properties of channels, Table IV 

of I-beams, Table III 

of standard angles, Table V 

of various sections, Table II 
Punch press frame, design of, 205 
Puzzolan cement, 298 

Eadius of gyration, 42 
Rankine's formula for columns, 126, 127 
for combined bending and torsion, 

142 
Rate of applying load, effect on strength 

of cement, 301 

Rattler test of paving brick, 334 
Reactions of supports, 49, 52, 53, 54, 55, 

90, 101 

Rectangular plates, 186 
Rectangular shafts in torsion, 144 
Red cedar, strength of, 344 
Red fir, strength of, 344, 349, 350 
Reduction of area of cross section, 14, 271 
Reenforced concrete, 313 

adhesion of reenforcement, 314 
area of reenforcement, 314 
beams, 72, 316, 317, 318, 319, 320, 

321, 322 
corrosion of metal reenforcement, 

313 

object of reenforcement, 313 
Relation between stress components, 20 
between elastic constants, 30 
between shear and bending mo- 
ment, 55 



INDEX 



371 



Resilience, defined, 94 

of circular shafts, 143 

modulus of, 94, 274 
Result of straight-line law, 35 
Retaining walls, angle of repose, 245 

bearing power of piles, 246 

bearing power of soils, 245 

column footings, 251 

formulas for, 257 

maximum earth pressure, 253 

ordinary foundations, 249 

thickness of, 260 

Wellington's formula, 248 
Rigidity, torsional and flexural, 148 
Riveted joints, 171 
Rivet pitch, 172 
Rope, 356 

hemp, 360 

manila, 360, 361 

sisal, 361 

wire, 356, 358, 359, 360 
Rubber belting, 363 

Rupture, modulus of. See Modulus of 
rupture 

Safety, factors of, 16 
St. Venant's formula for combined bend- 
ing and torsion, 142 
Sandstone, formation and properties of, 
327 

strength of, 328, 329 
Sapwood and heartwood, 337 
Seconds, timber, 349 
Section modulus, 40, Tables VI and VII 
Selects, timber, 349 
Setting, time of (cement), 299, 300 
Shear, and bending moment, relation 
between, 55 

and moment diagrams, Table XI, 
80 

at neutral axis, 324 

defined, 3 

influence line for, 98 

maximum, 25 

modulus of elasticity of, Table I 

simple, 27 

vertical reactions and, 49 
Shearing deflection, general formula 

for, 114 
Shearing strength of materials, average 

values, Table I 
Shearing tests, 342, 344, 346 
Shrinkage and forced fits, 168 
Silicon in iron and steel, 280, 289 
Simple shear, 27 
Sisal rope, 361 

Size, effect of size, of test piece on 
strength of steel, 272 

of timber, 350 
Slag cement, 298 



Slippery elm, strength of, 344 
Soundness test for cements, 298 
Sour gum, strength of, 344 
Specifications, for cast iron, 285 

for cement, 303 

for steel, 294 

for wrought iron, 294 
Speed of application of load, effect on 

strength of cement, 301 
Spheres and cylinders, 154, 155 
Springs, helical, 145 

general theory of spiral, 147 

hooks and links, 191 

plane spiral, 203 
Spruce, strength of, 346 
Square plates, 185 
Square shafts in torsion, 144 
Stability, of abutments, 229 

of arches, 220 

of retaining walls, 258 

Winkler's criterion for, 225 
Standard forms of test specimens, 294 
Steel, Bessemer process, 288 

castings, 291, 292, 293 

composition of, 288 

impurities in, 287, 288, 289 

manufacture and properties, 287 

modulus of elasticity of, 293 

nickel, 290 

open-hearth process, 288 

specifications for, 294 

strength of, 289, 290 

vanadium, 291 

Steel and concrete. See Reenforced con- 
crete 

Stone. See Limestone and Sandstone 
Straight-line formula, 320 
Straight-line law, 35 
Strain, defined, 2 

diagrams, 4, 281, 283, 311, 341 

equivalent, 31 

measure of, 31 

Strength, of materials, average values, 
Table I 

of reenforced concrete beams, 317 

of T-beams, 322 
Stress, defined, 2 

ellipse, 26 

maximum normal, 23 

trajectories, 71 
Stresses, equivalent, 31 

in different directions, 22 

temperature, 12, 236, 242 
Structure of timber, 336 
Struts and columns, 120 
Sugar maple, strength of, 344 
Sulphur in iron and steel, 280, 290 
Sweet gum, strength of, 344 
Sycamore, strength of, 344 
System of equivalent forces, 33 



372 



STRENGTH OF MATERIALS 



Tamarack, strength of, 344 
Temperature, effect on strength of steel, 
272 

effect of, on strength of concrete, 312 
Temperature stresses, 12, 236, 242 
Tensile strength, average values, 18 
Tensile tests, 266 

of belting, 362, 363 

of cast iron, 280, 286 

of cement, 301, 302 

of rope, 361 

of steel, 289, 290, 292 

of timber, 343 

of wire rope, 358, 359, 360 

of wrought iron, 294 
Tension, denned, 2 
Tenth census timber tests, 344 
Test pieces, large and small, 272, 350 
Test specimens, standard forms of, 

294 

Theorem of three moments, 90 
Theorems on moment of inertia, 40 
Theory of flat plates, 179 
Theory of flexure, common, 36 
Thick hollow cylinders. See Cylinders 
Thickness of retaining walls, 260 
Thin hollow cylinders. See Cylinders 
Timber, absorption process, 352 

annual rings, 336 

full-cell creosoting process, 352 

moisture in, 337 

results of tests, 344, 346, 347, 349, 
350 

Riiping process, 353 

sapwood and heartwood, 337 

strength of, 337 

strength of treated, 353 

structure of, 336 

tests of treated ties, 354 

treated, 351 

zinc chloride process, 352 
Time effect, 9 
Tool steel, 287 
Torsion, and bending combined, 32, 141 

angle of twist, 139, 145, 274 



Torsion, as test for shear, 277 

circular shafts in, 138, 139, 140 

elliptical shafts in, 144 

maximum stress in circular shafts 

in, 138 

non-circular shafts in, 143 
rectangular and square shafts in, 

144 

resilience of circular shafts in, 143 
test specimen, 276 
tests, 275 

triangular shafts in, 145 
Torsional rigidity, 148 
Transverse tests. See Flexure 
Tubes, collapse of, under external 

pressure, 159 

practical formulas for collapse of, 
166 

Ultimate strength, defined, 6, 7 
average values for, Table I 
Unit deformation, 4 
Unit stress, denned, 3 
United States Forest Service, 348 

Vanadium steel, 291 

Vertical reaction and shear, 49 

Wellington's formula, 248 

Winkler's criterion for stability, 225 

Wire, 356 

Wire rope. See Rope 

Wood. See Timber 

Work, defined. See Resilience 

of deformation of beams, 93 
Working stress in concrete beams, 322 
Wrought iron, impurities in, 290 

manufacture and properties, 286, 
290 

modulus of elasticity of, 293 

specifications for, 294 

Yield point, defined, 6, 270 
Young's modulus, 6 

average values for, Table I 



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